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(A) Let the number of stones be $N$.According to the problem,$N$ is a multiple of $21$,so $N = 21k$ for some integer $k$.Also,when $N$ is divided by $

Vedclass · Accepted Answer

$7203$

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(A) Let the number of stones be $N$.According to the problem,$N$ is a multiple of $21$,so $N = 21k$ for some integer $k$.Also,when $N$ is divided by $16, 20, 25,$ and $45$,the remainder is $3$ in each case.This means $N - 3$ must be exactly divisible by the $L.C.M.$ of $16, 20, 25,$ and $45$.First,find the $L.C.M.$ of $16, 20, 25, 45$:$16 = 2^4$$20 = 2^2 	imes 5$$25 = 5^2$$45 = 3^2 	imes 5$$L.C.M. = 2^4 	imes 3^2 	imes 5^2 = 16 	imes 9 	imes 25 = 3600$.So,$N = 3600m + 3$ for some integer $m$.We test values of $m$ such that $3600m + 3$ is divisible by $21$:For $m = 1$,$N = 3600(1) + 3 = 3603$. $3603 / 21 = 171.57$ (Not divisible).For $m = 2$,$N = 3600(2) + 3 = 7203$. $7203 / 21 = 343$ (Divisible).Thus,the minimum number of stones is $7203$.

$A$ heap of stones can be made up into groups of $21$. When made up into groups of $16, 20, 25$ and $45$,there are $3$ stones left in each case. How many stones at least can there be in the heap?

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