A positive integer is divisible by 9 if and only if the sum of its

Question

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*10^n + k*10^{2n}  divisible by 9?

(A) 9 
(B) 16 
(C) 23 
(D) 35
(E) 47

ScottTargetTestPrep · Accepted Answer

We need to determine for which value of k 25*10^n + k*10^2n is divisible by 9.

We see that 10^n and 10^2n will always have a digit of 1 and then zeros. So, excluding k, the sum of the digits in our expression is 2 + 5 = 7 (since (25)(10^n) has a 2, 5, and zeros).

We need to determine, of our answer choices, which when added to 7 will produce a sum that is divisible by 9. Scanning our answer choices, we see that 47 is the correct answer.

2 + 5 + 4 + 7 = 18, which is divisible by 9.

Answer: E

marcelonac · Answer

For any value of N, when you multiply by 10ˆn or 10ˆ2n you will be only adding zeros. What you only need to do is check if the sum of 2+5+the other possible values for k in the question add up to 9. The only possible answer is E.

racingip · Answer

Here is my approach:

It tells us that 25 × 10^n + k × 10^(2n) is divisible by 9 and we know that 10^whatever is not divisible by 9. So we just plug in numbers to find a number that satisfies that the sum of its digits (25 + k) is divisible by 9. 
Starting with option E: 25 + 47 = 72 and 7 + 2 = 9. Hence, when k=47 the number is divisible by 9.

vitaliyGMAT · Answer

25 = 7 (mod 9) = -2 (mod 9)

10^n = 10^{2n}  = 1 (mod 9)

-2 * 1 + k*1 = k - 2

k should have remainder 2 when divided by 9 to give us total remainder 0.

Only option which leaves remainder 2 upon division by 9 is 47 (E).

Ndkms · Answer

This looks like >620 imho

Solution:

25*10^n + k*10^2n = (25 * 10^n) + (k* 10^n * 10^n) {you should be able to see that 10^2n breaks into 10^n * 10^n else work your exponents roots in algebra}

then you can do 10^n *   And here the logic/critical thinking begins

a) The only thing that the very first 10^n does to the number within the square brackets is simply padding it with zero at the end. (as other users suggested before). So it doesn't play any role to the divisibility hence can be fully ignored (N.B. n>0).

b) now 25 + (k * 10^n) simply is K * mul(10) + 25 therefore you care only about the sum of the digits K, 2 and 5. By using "back-solving" you can find E

Therefore E.

good luck!

daviddaviddavid · Answer

I'm just using ridiculous numbers for these questions

so let n be= 10000000000000000

then we get something like:

250000000000.....+ k 00000000000000000000........

from this it becomes clear that we just have to look for 2+5 +  k  divisible by 9

7+47=54 which is divisible by 9

jkolachi · Answer

A very simple approach.

s8z8p9KU_xQ

Vegita · Answer

Hi  ScottTargetTestPrep

I did 25 + 47 = 72 and the sum 7+2 = 9

I don't understand how your solution worked 2 + 5 + 4 + 7 = 18. Since we have to get the actual value of dividend first and make sure the sum of the dividend is divisible by 9.

woohoo921 · Answer

ScottTargetTestPrep   MartyTargetTestPrep 
I realize that multiplying 25 by 10 to a power and so on for the k term does not impact the digits (besides adding zeroes.) However, I was confused on the rule itself defined by the equation, as I was unsure if I needed to add two ones (one from 10^n and one from 10^2n). So, I initially did 2+5+1+1=9. So, I chose A. How do you know you can't do this? Thank you!

MartyTargetTestPrep · Answer

Hi woohoo921.

The question could be a little clearer. It's true that it doesn't directly state that we are to find the sum of the digits of a single positive integer that is equal to 25*10^n + k*10^2n.

However, since the question asks about "A positive integer," and since the rule "A positive integer is divisible by 9 if and only if the sum of its digits is divisible by 9" doesn't apply to multiple numbers within an equation or expression but applies only to "a positive integer," we can reason that the question is about a single positive integer, and not about 25, 10, k and 10 within the mathematical expression.

Thus, we can tell that we need to consider the integer that is the result of 25*10^n + k*10^2n to find the answer, and for that reason the ones associated with the tens don't matter.

captainPirque · Answer

Hi, could someone please explain this as I am stuck with the same question?

Bunuel · Answer

To verify whether a number is divisible by 9, we can use the rule that states that the sum of its digits must be divisible by 9. For example, to check if 72 is divisible by 9, we can add its digits (7+2 = 9) and check if the result is divisible by 9. Since 9 is indeed divisible by 9, we can conclude that 72 is also divisible by 9.

This rule can be extended to verify if the sum of any set of integers is divisible by 9. We can simply add all the integers in the set and check if the resulting sum is divisible by 9. Alternatively, we can add the digits of each integer in the set and check if the sum of these digits is divisible by 9. For example, to verify if 40 + 32 is divisible by 9, we can add the digits of each integer (4+0+3+2 = 9) and check if the result is divisible by 9. Since 9 is indeed divisible by 9, we can conclude that 40 + 32 is also divisible by 9.

For the given problem, we can check if  25*10^n + k*10^{2n}  is divisible by 9 by either finding the sum of its digits and verifying if it is divisible by 9, or by calculating the sum of the digits of its two components,  25*10^n  and  k*10^{2n} , separately, and checking if their sum is divisible by 9. Both methods will give the same result.

The sum of the digits of  25*10^n  is always equal to 2 + 5 = 7, regardless of the value of n (assuming n is a positive integer). Similarly, the sum of the digits of  k*10^{2n}  is equal to the sum of the digits of k, again regardless of the value of n (assuming n is a positive integer).

Therefore, to find whether  25*10^n + k*10^{2n}  is divisible by 9, we can add 7 to the sum of the digits of k and check if the resulting sum is divisible by 9. Among the answer choices, only 47 (option E) satisfies this condition, where the sum of its digits (4+7) added to 7 gives a total sum that of 18, which is divisible by 9.

Hope it's clear.

Vegita · Answer

Bunuel  Thank you for the clarification!

captainPirque · Answer

Bunuel  thank you for the clarification!

Does the same logic apply to divisibility by 3?

Bunuel · Answer

___________________________________________
Absolutely!

gmattopu · Answer

This is not always the case, for example if you take n = 1.

Bunuel · Answer

If n = 1 and k = 47, then  25*10^1 + 47*10^2  equals 4950, which is divisible by 9, just as all other values of  25*10^n + k*10^{2n}  are, where n is a positive integer and k = 47. Please review the discussion above carefully for a better understanding. I hope it helps.

ManifestDreamMBA · Answer

Sum of digits should be 9
we have 2+5 (rest are all 0s) = 7
we need a number whose digits add up to 2. E works

A positive integer is divisible by 9 if and only if the sum of its

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