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A positive integer is divisible by 9 if and only if the sum of its digits is divisible by 9 . If n is a positive integer, for which of the following values of k is 25 \times 1 0^{n} + k \times 1 0^{2 n} divisible by 9 ?

Explanation	Calculations	Help
$We apply the divisibility rule for 9, which states that a number is divisible by 9 exactly when the sum of its digits is a multiple of 9.$	$A number is divisible by 9 ⟺ sum of its digits is a multiple of 9$	Theory & Tactics Method Card DIV3-C Click to view full details
$We find the digit sum of the term 25 \times 1 0^{n} . Appending zeros does not change the nonzero digits.$	$sum of digits (25 \times 1 0^{n}) = 2 + 5 = 7$
$We find the digit sum of the term k \times 1 0^{2 n} . Appending zeros does not change the digit sum of k .$	$sum of digits (k \times 1 0^{2 n}) = sum of digits (k)$
$We add these digit sums to get the total digit sum of the entire expression.$	$sum of digits (25 \times 1 0^{n} + k \times 1 0^{2 n}) = 7 + sum of digits (k)$
$For divisibility by 9, the total must be a multiple of 9. We test each choice by computing 7 + sum of digits (k) .$	$7 + sum of digits (9) = 7 + 9 = 16 7 + sum of digits (16) = 7 + (1 + 6) = 14 7 + sum of digits (23) = 7 + (2 + 3) = 12 7 + sum of digits (35) = 7 + (3 + 5) = 15 7 + sum of digits (47) = 7 + (4 + 7) = 18$
$We see that only for the choice 47 the total digit sum is a multiple of 9, so we select 47.$	$Answer = 47$

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47

A positive integer is divisible by 9 if and only if the sum of its digits is divisible by 9 . If n is a positive integer, for which of the following values of k is 25 \times 1 0^{n} + k \times 1 0^{2 n} divisible by 9 ?

Explanation	Calculations	Help
$We apply the divisibility rule for 9, which states that a number is divisible by 9 exactly when the sum of its digits is a multiple of 9.$	$A number is divisible by 9 ⟺ sum of its digits is a multiple of 9$	Theory & Tactics Method Card DIV3-C Click to view full details
$We find the digit sum of the term 25 \times 1 0^{n} . Appending zeros does not change the nonzero digits.$	$sum of digits (25 \times 1 0^{n}) = 2 + 5 = 7$
$We find the digit sum of the term k \times 1 0^{2 n} . Appending zeros does not change the digit sum of k .$	$sum of digits (k \times 1 0^{2 n}) = sum of digits (k)$
$We add these digit sums to get the total digit sum of the entire expression.$	$sum of digits (25 \times 1 0^{n} + k \times 1 0^{2 n}) = 7 + sum of digits (k)$
$For divisibility by 9, the total must be a multiple of 9. We test each choice by computing 7 + sum of digits (k) .$	$7 + sum of digits (9) = 7 + 9 = 16 7 + sum of digits (16) = 7 + (1 + 6) = 14 7 + sum of digits (23) = 7 + (2 + 3) = 12 7 + sum of digits (35) = 7 + (3 + 5) = 15 7 + sum of digits (47) = 7 + (4 + 7) = 18$
$We see that only for the choice 47 the total digit sum is a multiple of 9, so we select 47.$	$Answer = 47$

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47