4767 exactly divides xxx,341. The missing dig | vedclass

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Question

(B) Let the number be $N = d_2 d_1 d_0 341$,where $d_2, d_1, d_0$ are the missing digits.Since $4767$ divides $N$ exactly,we can write $N = 4767 	ime

Vedclass · Accepted Answer

$586$

Vedclass · Answer

(B) Let the number be $N = d_2 d_1 d_0 341$,where $d_2, d_1, d_0$ are the missing digits.Since $4767$ divides $N$ exactly,we can write $N = 4767 	imes Q$.The last digit of $N$ is $1$ and the last digit of $4767$ is $7$.For the product to end in $1$,the last digit of the quotient $Q$ must be $3$ (since $7 	imes 3 = 21$).Let us estimate the quotient $Q$ by dividing $xxx,341$ by $4767$.$4767 	imes 100 = 476700$.$N - 476700 = xxx,341 - 476,700 = yyy,641$.Now,$4767 	imes 20 = 95340$.$N - 476700 - 95340 = 586341 - 572040 = 14301$.Finally,$4767 	imes 3 = 14301$.Thus,$Q = 100 + 20 + 3 = 123$.$N = 4767 	imes 123 = 586341$.The missing digits are $5, 8, 6$.

$4767$ exactly divides $xxx\,341$. The missing digits are:

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