Fundamental Theorem of Arithmetic Questions in English

(A) Euclid's division lemma states that for any two positive integers $a$ and $b,$ there exist unique integers $q$ and $r$ such that $a = bq + r,$ where the remainder $r$ must satisfy the condition $0 \leq r < b.$ This means the remainder $r$ is greater than or equal to $0$ and strictly less than the divisor $b.$ Therefore,the correct condition is $0 \leq r < b.$

(B) Euclid's division lemma states that for any two positive integers $a$ and $b$,there exist unique integers $q$ and $r$ such that $a = bq + r$,where $0 \leqslant r < b$.
In this problem,we are given $b = 7$.
Substituting $b = 7$ into the condition $0 \leqslant r < b$,we get $0 \leqslant r < 7$.
Therefore,the correct condition for the remainder $r$ is $0 \leqslant r < 7$.

(D) According to Euclid's division lemma,for any two positive integers $a$ and $b$,there exist unique integers $q$ and $r$ such that $a = bq + r$,where $0 \le r < b$.
In this question,$b = 5$.
Therefore,the possible values for $r$ are $0, 1, 2, 3, 4$.
Since $r$ must be less than $5$,the value $r = 6$ is not possible.

(A) We are given the expression $2^{m} \cdot 5^{n}$ where $m, n \in N$ (natural numbers).
We can rewrite the expression as $2^{m} \cdot 5^{n} = 2^{m-n} \cdot 2^{n} \cdot 5^{n}$ (if $m \ge n$) or $2^{m} \cdot 5^{m} \cdot 5^{n-m}$ (if $n > m$).
Case $1$: If $m = n$,then $2^{m} \cdot 5^{m} = (2 \cdot 5)^{m} = 10^{m}$. The last digit of $10^{m}$ is $0$.
Case $2$: If $m > n$,then $2^{m} \cdot 5^{n} = 2^{m-n} \cdot (2 \cdot 5)^{n} = 2^{m-n} \cdot 10^{n}$. Since $2^{m-n}$ is an even number and $10^{n}$ ends in $0$,the product will end in $0$.
Case $3$: If $n > m$,then $2^{m} \cdot 5^{n} = (2 \cdot 5)^{m} \cdot 5^{n-m} = 10^{m} \cdot 5^{n-m}$. Since $5^{n-m}$ always ends in $5$ for $n > m$ and $10^{m}$ ends in $0$,the product will end in $0$.
In all cases,the last digit is $0$.

(C) To find the number of decimal places for a rational number $\frac{p}{q}$ where $q = 2^n \cdot 5^m$,the number of decimal places is given by $\max(n, m)$.
Given the expression $\frac{18}{5^3}$,we can rewrite the denominator as $2^0 \cdot 5^3$.
Here,$n = 0$ and $m = 3$.
The number of decimal places is $\max(0, 3) = 3$.
Alternatively,$\frac{18}{5^3} = \frac{18 \cdot 2^3}{5^3 \cdot 2^3} = \frac{18 \cdot 8}{10^3} = \frac{144}{1000} = 0.144$.
Since there are $3$ digits after the decimal point,the correct option is $C$.

(C) For any natural number $n \in N$ (where $n = 1, 2, 3, \dots$):
If $n = 1$,$5^1 = 5$.
If $n = 2$,$5^2 = 25$.
If $n = 3$,$5^3 = 125$.
If $n = 4$,$5^4 = 625$.
It is observed that for any positive integer $n$,the last digit of $5^n$ is always $5$.

(C) To find the last digit of $5^n$ for any positive integer $n$,let us observe the powers of $5$:
$5^1 = 5$
$5^2 = 25$
$5^3 = 125$
$5^4 = 625$
In each case,the last digit (units digit) is $5$.
Therefore,for any positive integer $n$,the last digit of $5^n$ is always $5$.

(A) We are given the expression $2^5 \cdot 5^5$.
Using the law of exponents $a^n \cdot b^n = (a \cdot b)^n$,we can rewrite the expression as:
$2^5 \cdot 5^5 = (2 \cdot 5)^5$
$= 10^5$
$= 100000$
The last digit of $100000$ is $0$.

(C) According to the Fundamental Theorem of Arithmetic,every composite number greater than $1$ can be expressed as a product of prime numbers. This representation is unique,apart from the order in which the prime factors occur. This theorem is also known as the Unique Factorization Theorem.

(B) To find the last digit of $5^{25}$,we observe the powers of $5$:
$5^1 = 5$
$5^2 = 25$
$5^3 = 125$
$5^4 = 625$
It is observed that for any positive integer $n$,the last digit of $5^n$ is always $5$.
Therefore,the last digit of $5^{25}$ is $5$.

(C) To find the last digit of $5^n$ for any positive integer $n$,let us observe the powers of $5$:
$5^1 = 5$
$5^2 = 25$
$5^3 = 125$
$5^4 = 625$
As we can see,for any positive integer $n$,the result of $5^n$ always ends in the digit $5$. Therefore,the last digit is always $5$.

(C) To find the prime factorization of $30$,we divide it by the smallest prime numbers:
$30 \div 2 = 15$
$15 \div 3 = 5$
$5 \div 5 = 1$
Therefore,the prime factorization of $30$ is $2 \times 3 \times 5$.

(A) To find the remainder when $(5k+1)^2$ is divided by $5$,we expand the expression using the identity $(a+b)^2 = a^2 + 2ab + b^2$.
$(5k+1)^2 = (5k)^2 + 2(5k)(1) + (1)^2$
$= 25k^2 + 10k + 1$
Now,we can rewrite this expression as:
$= 5(5k^2 + 2k) + 1$
Since $5(5k^2 + 2k)$ is clearly divisible by $5$,the remainder when the entire expression is divided by $5$ is $1$.

(B) To find the last digit of $5^{25}$,we observe the powers of $5$:
$5^1 = 5$
$5^2 = 25$
$5^3 = 125$
$5^4 = 625$
Since the last digit of any positive integer power of $5$ is always $5$,the last digit of $5^{25}$ is $5$.

(C) To find the last digit of $5^n$ for any positive integer $n$:
$5^1 = 5$
$5^2 = 25$
$5^3 = 125$
$5^4 = 625$
As observed,for any positive integer $n$,the product of $5$ multiplied by itself $n$ times will always result in a number ending in $5$. Therefore,the last digit of $5^n$ is always $5$.

(A) Let the three consecutive positive integers be $n, (n+1)$,and $(n+2)$.
The product of these integers is $P = n(n+1)(n+2)$.
This expression represents the product of three consecutive integers.
We know that in any set of $k$ consecutive integers,there is always at least one multiple of $k!$ (k factorial).
For $k=3$,the product is always divisible by $3! = 3 \times 2 \times 1 = 6$.
For example,if $n=1$,$1 \times 2 \times 3 = 6$ (divisible by $6$).
If $n=2$,$2 \times 3 \times 4 = 24$ (divisible by $6$).
If $n=3$,$3 \times 4 \times 5 = 60$ (divisible by $6$).
Thus,the product of any three consecutive positive integers is always divisible by $6$.

(C) To find the prime factorization of $30$,we divide it by the smallest prime numbers:
$30 = 2 \times 15$
$15 = 3 \times 5$
Therefore,$30 = 2 \times 3 \times 5$.
Since $2, 3,$ and $5$ are all prime numbers,the correct prime factorization is $2 \times 3 \times 5$.

(C) According to the Fundamental Theorem of Arithmetic,every composite number greater than $1$ can be uniquely expressed as a product of primes,except for the order in which the prime factors occur. This means that for any given composite number,the set of prime factors is fixed,regardless of the order in which they are written. Therefore,the factorization is unique.

(C) To find the number of decimal places,we express the denominator in the form $2^n \times 5^m$.
Given fraction: $\frac{35457}{6250}$.
Prime factorization of $6250 = 625 \times 10 = 5^4 \times (2 \times 5) = 2^1 \times 5^5$.
The number of decimal places is determined by the maximum of the exponents of $2$ and $5$,which is $\max(1, 5) = 5$.
Therefore,the decimal representation terminates after $5$ decimal places.

(C) By Euclid's division lemma,for any positive integer $a$ and divisor $b=4$,we have $a = 4m + r$,where $r$ can be $0, 1, 2, 3$.
If $r=0$,$a = 4m = 2(2m)$,which is even.
If $r=1$,$a = 4m+1$,which is odd.
If $r=2$,$a = 4m+2 = 2(2m+1)$,which is even.
If $r=3$,$a = 4m+3$,which is odd.
Therefore,any odd positive integer is of the form $4m+1$ or $4m+3$.

(C) Euclid's division lemma states that for any two positive integers $a$ and $b$,there exist unique integers $q$ and $r$ such that $a = bq + r$,where $0 \le r < b$.
In this problem,$b = 5$.
Therefore,the possible values for the remainder $r$ are $0, 1, 2, 3, 4$.
Since $r$ must be less than $5$,the value $r = 6$ is not possible.

(B) Any natural number $a$ can be expressed in the form $6k, 6k+1, 6k+2, 6k+3, 6k+4,$ or $6k+5$.
Calculating $a^2 \pmod{6}$ for each case:
$1) (6k)^2 = 36k^2 = 6(6k^2) \equiv 0 \pmod{6}$
$2) (6k+1)^2 = 36k^2 + 12k + 1 = 6(6k^2 + 2k) + 1 \equiv 1 \pmod{6}$
$3) (6k+2)^2 = 36k^2 + 24k + 4 = 6(6k^2 + 4k) + 4 \equiv 4 \pmod{6}$
$4) (6k+3)^2 = 36k^2 + 36k + 9 = 6(6k^2 + 6k + 1) + 3 \equiv 3 \pmod{6}$
$5) (6k+4)^2 = 36k^2 + 48k + 16 = 6(6k^2 + 8k + 2) + 4 \equiv 4 \pmod{6}$
$6) (6k+5)^2 = 36k^2 + 60k + 25 = 6(6k^2 + 10k + 4) + 1 \equiv 1 \pmod{6}$
The possible remainders are $0, 1, 3, 4$.
Thus,the remainders $2$ and $5$ are not possible.

(C) To find the last digit of $5^n$ for any positive integer $n$,let's observe the powers of $5$:
$5^1 = 5$
$5^2 = 25$
$5^3 = 125$
$5^4 = 625$
As we can see,for any positive integer $n$,the last digit of $5^n$ is always $5$.

(B) To find the number of decimal places for a rational number $\frac{p}{q}$ where $q = 2^n \cdot 5^m$,the number of decimal places is equal to $\max(n, m)$.
Given the fraction is $\frac{15}{2^2}$.
Here,the denominator is $2^2 \cdot 5^0$.
Comparing this with $2^n \cdot 5^m$,we get $n = 2$ and $m = 0$.
The maximum value of $n$ and $m$ is $\max(2, 0) = 2$.
Therefore,the decimal expansion terminates after $2$ decimal places.
Verification: $\frac{15}{2^2} = \frac{15}{4} = 3.75$,which has $2$ decimal places.

(B) We are given the expression $(6k+1)^2$.
Expanding this using the algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$,we get:
$(6k+1)^2 = (6k)^2 + 2(6k)(1) + 1^2$
$= 36k^2 + 12k + 1$
We can factor out $6$ from the first two terms:
$= 6(6k^2 + 2k) + 1$
Let $m = 6k^2 + 2k$,where $m$ is an integer.
Then the expression becomes $6m + 1$.
When $6m + 1$ is divided by $6$,the quotient is $m$ and the remainder is $1$.

(A) To determine if a rational number $p/q$ has a terminating decimal expansion,we check the prime factorization of the denominator $q$. If $q$ is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers,then the decimal expansion is terminating.
Here,the fraction is $44/625$.
The denominator is $625 = 5^4$.
Since the denominator is in the form $2^0 \times 5^4$,it satisfies the condition for a terminating decimal expansion.
Therefore,$44/625$ is a terminating decimal.

(A) We are given the expression $2^5 \cdot 5^5$.
Using the law of exponents $a^n \cdot b^n = (a \cdot b)^n$,we can rewrite the expression as:
$2^5 \cdot 5^5 = (2 \cdot 5)^5$
$= 10^5$
$= 100,000$
The last digit of $100,000$ is $0$.

(C) For any natural number $n \in N$,the expression $5^n$ represents the product of $5$ multiplied by itself $n$ times.
For $n = 1$,$5^1 = 5$.
For $n = 2$,$5^2 = 25$.
For $n = 3$,$5^3 = 125$.
For $n = 4$,$5^4 = 625$.
As observed,the last digit of $5^n$ is always $5$ for any natural number $n$.

(C) To find the number of decimal places for a rational number $\frac{p}{q}$ where $q = 2^n \cdot 5^m$,the number of decimal places is equal to $\max(n, m)$.
Given the expression $\frac{18}{5^{3}}$,we can rewrite the denominator as $2^0 \cdot 5^3$.
Here,$n = 0$ and $m = 3$.
The number of decimal places is $\max(0, 3) = 3$.
Thus,the decimal expansion terminates after $3$ decimal places.

(B) Euclid's division lemma states that for any two positive integers $a$ and $b$,there exist unique integers $q$ and $r$ such that $a = bq + r$,where the remainder $r$ satisfies the condition $0 \leq r < b$. This means the remainder $r$ can be zero or greater than zero,but it must always be strictly less than the divisor $b$.

(A) We are given the expression $2^{m} \cdot 5^{n}$ where $m, n \in N$.
We can rewrite this as $2^{m-n} \cdot (2 \cdot 5)^{n}$ if $m \ge n$,or $5^{n-m} \cdot (2 \cdot 5)^{m}$ if $n > m$.
In either case,the expression contains a factor of $(2 \cdot 5) = 10$.
Any positive integer multiplied by $10$ will result in a number ending in $0$.
For example,if $m=1, n=1$,then $2^1 \cdot 5^1 = 10$ (ends in $0$).
If $m=2, n=1$,then $2^2 \cdot 5^1 = 4 \cdot 5 = 20$ (ends in $0$).
If $m=1, n=2$,then $2^1 \cdot 5^2 = 2 \cdot 25 = 50$ (ends in $0$).
Therefore,the last digit is always $0$.

(C) To convert the fraction $\frac{1}{32}$ into decimal form,we perform long division.
Divide $1$ by $32$:
$1 \div 32 = 0.03125$.
Alternatively,we can write $\frac{1}{32} = \frac{1}{2^5}$.
To convert to a decimal,multiply the numerator and denominator by $5^5$:
$\frac{1 \times 5^5}{2^5 \times 5^5} = \frac{3125}{10^5} = \frac{3125}{100000} = 0.03125$.
Thus,the correct option is $C$.

(B) rational number $\frac{p}{q}$ has a terminating decimal expansion if and only if the prime factorization of the denominator $q$ is of the form $2^m \cdot 5^n$,where $m$ and $n$ are non-negative integers.
Non-negative integers include all natural numbers $(N)$ and zero $(0)$.
Therefore,$m, n \in N \cup \{0\}$.

(C) To find the number of decimal places after which the decimal expansion of $\frac{2517}{6250}$ terminates,we need to express the denominator in the form $2^n \times 5^m$.
First,find the prime factorization of the denominator $6250$:
$6250 = 625 \times 10 = 5^4 \times (2 \times 5) = 2^1 \times 5^5$.
The decimal expansion of a rational number $\frac{p}{q}$ terminates after $k$ decimal places,where $k = \max(n, m)$ in the prime factorization $q = 2^n \times 5^m$.
Here,$n = 1$ and $m = 5$.
Therefore,$k = \max(1, 5) = 5$.
The decimal expansion will terminate after $5$ decimal places.

(B) rational number $\frac{p}{q}$ has a terminating decimal expansion if the prime factorization of the denominator $q$ is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers.
$1$. For $\frac{17}{32}$: $32 = 2^5$. Since it is in the form $2^n \times 5^m$,it has a terminating decimal expansion.
$2$. For $\frac{17}{248}$: $248 = 8 \times 31 = 2^3 \times 31^1$. Since there is a factor other than $2$ or $5$ (i.e.,$31$),it has a non-terminating repeating decimal expansion.
$3$. For $\frac{17}{160}$: $160 = 16 \times 10 = 2^4 \times 2 \times 5 = 2^5 \times 5^1$. Since it is in the form $2^n \times 5^m$,it has a terminating decimal expansion.
Therefore,$\frac{17}{248}$ has a non-terminating repeating decimal expansion.

(D) To find the number of decimal places in the decimal representation of a rational number $\frac{p}{q}$ (where $q \neq 0$),we first simplify the fraction to its lowest terms.
$\frac{7}{80}$ is already in its simplest form.
Next,we find the prime factorization of the denominator $q = 80$.
$80 = 2^4 \times 5^1$.
The number of decimal places is given by the maximum exponent of $2$ or $5$ in the prime factorization of the denominator.
Here,the exponents are $4$ and $1$. The maximum exponent is $4$.
Therefore,the decimal representation of $\frac{7}{80}$ will have $4$ digits after the decimal point.
Verification: $\frac{7}{80} = \frac{7 \times 125}{80 \times 125} = \frac{875}{10000} = 0.0875$.

(D) In a factor tree, a number is represented as the product of its factors.
Using the property: $\text{Number} = \text{Factor}_1 \times \text{Factor}_2$.
For the first branch, $x$ is the product of $2$ and $5$, so $x = 2 \times 5 = 10$.
For the second branch, $y$ is the product of $x$ and $2$, so $y = x \times 2 = 10 \times 2 = 20$.
Thus, the values are $x = 10$ and $y = 20$.

(B) By Euclid's division lemma,any positive integer $a$ can be expressed in the form $a = bq + r$,where $0 \le r < b$.
For $b = 2$,the possible values of $r$ are $0$ and $1$.
Thus,$a = 2m + 0 = 2m$ (which is an even integer) or $a = 2m + 1$ (which is an odd integer).
Therefore,any odd positive integer is of the form $2m + 1$.

(C) Euclid's division lemma states that for any two positive integers $a$ and $b$,there exist unique integers $q$ and $r$ such that $a = bq + r$,where $0 \le r < b$.
In this problem,$b = 5$.
Therefore,the condition for $r$ is $0 \le r < 5$.
This means the possible values for $r$ are ${0, 1, 2, 3, 4}$.
Comparing this with the given options,$5$ is not a possible value for $r$ because $r$ must be strictly less than the divisor $5$.

(A) We are given the expression $2^{3} \times 5^{125}$.
First,calculate $2^{3} = 8$.
Now,the expression becomes $8 \times 5^{125}$.
We can rewrite this as $8 \times 5 \times 5^{124} = 40 \times 5^{124}$.
Since any number multiplied by $40$ ends in $0$,the last digit of the expression is $0$.

(B) To determine the nature of the number,we simplify the expression:
$3 \times 15 \times 11 + 11$
$= 11 \times (3 \times 15 + 1)$
$= 11 \times (45 + 1)$
$= 11 \times 46$
$= 11 \times 2 \times 23$
Since the number can be expressed as a product of prime factors other than $1$ and itself (i.e.,$2, 11, 23$),it is a composite number.

(A) Let the positive integer be $x = 3q + 1$.
To find the square of this integer,we calculate $x^2 = (3q + 1)^2$.
Using the algebraic identity $(a + b)^2 = a^2 + 2ab + b^2$,we get:
$x^2 = (3q)^2 + 2(3q)(1) + (1)^2$
$x^2 = 9q^2 + 6q + 1$
We can factor out $3$ from the first two terms:
$x^2 = 3(3q^2 + 2q) + 1$
Let $m = 3q^2 + 2q$,where $m$ is an integer.
Thus,$x^2 = 3m + 1$.
Therefore,the statement is true.

(A) Let the three consecutive positive integers be $n, (n+1)$,and $(n+2)$.
In any set of three consecutive integers,at least one must be even (divisible by $2$) and exactly one must be a multiple of $3$.
Since the product contains at least one factor of $2$ and at least one factor of $3$,the product must be divisible by $2 \times 3 = 6$.
For example,if $n=1$,the product is $1 \times 2 \times 3 = 6$,which is divisible by $6$.
If $n=2$,the product is $2 \times 3 \times 4 = 24$,which is divisible by $6$.
Thus,the statement is true.

(A) Let the two consecutive positive integers be $n$ and $n+1$.
Their product is $P = n(n+1)$.
Since $n$ and $n+1$ are consecutive integers,one of them must be even and the other must be odd.
An even number is always divisible by $2$.
Therefore,the product of an even number and an odd number is always even,meaning it is divisible by $2$.
Thus,the statement is true.

(B) To find the number of digits after the decimal point for a rational number of the form $\frac{p}{2^n \cdot 5^m}$,we look for the maximum value between the exponents $n$ and $m$.
Given the expression $\frac{7}{2^3 \cdot 5^4}$,we have $n = 3$ and $m = 4$.
The number of decimal places is determined by $\max(n, m) = \max(3, 4) = 4$.
Thus,the decimal representation will have $4$ digits after the decimal point.

(A) To determine if the decimal expansion of a rational number $\frac{p}{q}$ is terminating,we check the prime factorization of the denominator $q$.
If $q$ is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers,the decimal expansion is terminating.
Here,$q = 3125$.
$3125 = 5 \times 5 \times 5 \times 5 \times 5 = 5^5$.
Since $3125$ can be written as $2^0 \times 5^5$,it satisfies the condition $2^n \times 5^m$.
Therefore,the decimal expansion of $\frac{13}{3125}$ is terminating.

(C) composite number is a positive integer that has at least one divisor other than $1$ and itself.
For the first expression: $7 \times 11 \times 13 + 13 = 13 \times (7 \times 11 + 1) = 13 \times (77 + 1) = 13 \times 78 = 13 \times 13 \times 6$.
Since this number can be expressed as a product of factors other than $1$ and itself,it is a composite number.
For the second expression: $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) = 5 \times (1008 + 1) = 5 \times 1009$.
Since $1009$ is a prime number,the expression is $5 \times 1009$. This number has factors $1, 5, 1009$,and the number itself. Since it has more than two factors,it is a composite number.

(B) For any number to end with the digit $0$,its prime factorization must contain the factors $2$ and $5$.
The prime factorization of $6^n$ is $(2 \times 3)^n = 2^n \times 3^n$.
Since the prime factorization of $6^n$ only contains $2$ and $3$,it does not contain the factor $5$.
Therefore,by the Fundamental Theorem of Arithmetic,$6^n$ cannot end with the digit $0$ for any natural number $n$.

(A) To find the prime factors of $7429$,we test divisibility by prime numbers sequentially.
$7429$ is not divisible by $2, 3, 5, 7, 11,$ or $13$.
Testing $17$: $7429 \div 17 = 437$.
Now,we find the prime factors of $437$.
Testing $19$: $437 \div 19 = 23$.
Since $23$ is a prime number,the prime factorization is $17 \times 19 \times 23$.

(A) To find the prime factors of $5005$,we perform successive division by prime numbers:
$1$. Since the last digit is $5$,it is divisible by $5$: $5005 \div 5 = 1001$.
$2$. Now,divide $1001$ by the next prime number,$7$: $1001 \div 7 = 143$.
$3$. Next,divide $143$ by the next prime number,$11$: $143 \div 11 = 13$.
$4$. Finally,$13$ is a prime number,so $13 \div 13 = 1$.
Therefore,the prime factorization of $5005$ is $5 \times 7 \times 11 \times 13$.