Find the nearest number to 12199 which is exa | vedclass

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Question

(A) The first four prime numbers are $2, 3, 5,$ and $7$.Their product is $2 	imes 3 	imes 5 	imes 7 = 210$.To find the nearest number to $12199$ di

Vedclass · Accepted Answer

$12180$

Vedclass · Answer

(A) The first four prime numbers are $2, 3, 5,$ and $7$.Their product is $2 	imes 3 	imes 5 	imes 7 = 210$.To find the nearest number to $12199$ divisible by $210$,we divide $12199$ by $210$:$12199 \div 210 = 58$ with a remainder of $19$.Since the remainder $19$ is less than half of the divisor $(210 / 2 = 105)$,the nearest multiple is obtained by subtracting the remainder from the dividend.Therefore,the required number is $12199 - 19 = 12180$.

Find the nearest number to $12199$ which is exactly divisible by the product of the first four prime numbers.

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