Define A = The sum of digits of the number 7^100 and B = The sum of

Shrouded1:

Read the digital sum concept ->> digital-sum-concept-100534.html

Digital sum of number is not 16 it will be reduced to 1+6= 7

Now you can infinitely reduced 07= 0+7 it will be = 7 only.

The question does not ask for what you define as "digital sum" it merely asks for sum of digits of B, which in turn is the sum of digits of A .....

Eg if A were 99999999999
B would be 99
and the answer would be 18 which is not the digital sum, only because we havent summed the digits enough number of times to get to the digital sum yet

Also just a side point, an easier way to think of digital sum is as follows :

Digital_sum(x) = 9 if (9 divides x); x mod 9 otherwise

Makes it much easier to manipulate mathematically with this definition

In the given question, since we are summing 3 times the digits, it is not necessary we arrive at the digital sum. So the anser need not be a single digit in base-10 representation, but the sum of its digits will eventually yield 7, i agree with that

Why you considered 99 = 9+ 9 = 18 ??? if the sum of digits is 99 you should not have reduced it to 18
Now if you have reduced it to 18, then why not to reduce 1+8 to 9


Shrouded1: When you are making multiple posts consecutively, consider editing the previous post.



Regarding the number 9, it is mentioned in the link that I have given above.
Don't you think you are contradicting yourself --> check the red part.

On one side you are saying it need not to be 7, it is not compulsory to reduce it to single digit, if that is the case then how 7 can be achieved ?

7 must have come using a+b = 7.

Apologies for the confusing post

All I am trying to say is that the digital sum is obtained by summing the digits of the number again and again till the final number you get is only a single digit one. In this case what we have done is sum the digits of the number thrice ... Now let's say that the number in question was not 7^100 but was 7^10000000, then it could well be the case that summing digits thrice is not enough to get to the digital sum of the number. What I do agree with 100% is that the digital sum is 7 and that of all intermediate numbers is also 7 but whether you get there within 3 operations of summing or whether it takes more needs to be proven separately and is a function of how big the starting number is

Posted from my mobile device

What level is this type of question considered?

Thanks.

I m not sure I got your point or not but if I can prove digital sum of 7^4 = 7

=> digital sum of \(7^{100,00,00,00)\)= digital sum of 7*7*7*7 = digital sum of 7^4 = 7

I think you should read the link given by me again.

Digital sum of a*b*c = digital sum of a* digital sum of b* digital sum of c

I guess 650-700.

Bunnel : Let me know if there is any issue with the above stated solutions.

Is this a GMAT question? If so, can I ask someone to provide a shorter explanation than the one by gurpreetsingh?

Also, is the concept of digital sum of the number and all the related properties useful for the GMAT?

Following approach will take you only couple of sec.

\(7^{100} = (7^{50})^2\)

The sum of digits (digital root) of a PERFECT SQUARE can only be 1, 4, 7 or 9.

For example \(7^2=49\) => DR=4+9=13 => 1+3=4 and etc.

Checking available options.

Answer A.

This question is not, as suggested above, a 650-700 level question. It's more like a 900-level problem, and it's not something you'd ever see on the GMAT. The solutions above are incomplete; they answer the question assuming we'll be summing digits until we get down to a single-digit number. That is what will end up happening here, but I don't think that's obvious just reading the question.

When you divide a number by 9, the remainder you get will always be equal to the remainder you get when you divide the sum of the number's digits by 9. So, for example, if you divide 3725 by 9, the remainder you'd get will be equal to the remainder you get when you divide 3+7+2+5 = 17 by 9, so it will be 8. That's a fairly straightforward consequence of the divisibility test for 9 (where you sum the digits to test if a number is divisible by 9), just adapted to remainders instead of divisibility.

If we now consider powers of 7, as in this question, the number 7^3 = 343 will give us a remainder of 1 when we divide by 9 (which you can see by summing its digits, or noticing 342 is a multiple of 9, and 343 = 342 + 1). If you know modular/remainder arithmetic (which is never tested in this way on the GMAT,) you'll know that we can 'multiply remainders' if we're always dividing by a fixed number, so if the remainder is 1 when we divide 7^3 by 9, the remainder will be 1^33 = 1 when we divide (7^3)^33 = 7^99 by 9, and when we divide 7^100 = (7)(7^99) by 9, the remainder will be (7)(1) = 7.

As a consequence of the second paragraph above, since the remainder is 7 when we divide 7^100 by 9, when we sum the digits of 7^100 (to get the number "A" described in the question), we must also get a number with a remainder of 7 when we divide by 9. There's no way to work out precisely what A is equal to here without a computer, but we do know we're adding the digits of 7^100, which is less than 10^100, a 101-digit number. So we're adding fewer than 101 digits, each of which will be at most 9, and the value of A cannot possibly be bigger than 900. When we add the digits of A to get the number called "B", we'll be adding three digits that are at most 9, so we'll get a number less than 27, and finally when we add the digits of this number to finally answer the question, we're adding a tens digit that is at most '2' to a units digit, so we must get something 11 or less. But we also must get something with a remainder of 7 when we divide by 9, and the only such number is 7 itself, so that must be the answer to the question.

You could technically answer the question more quickly just by estimating how large the answer can be (it needs to be very small because we add the digits of our numbers so many times, and only one answer choice is reasonable; note that E is clearly wrong since the answer is obviously calculable somehow) but then you're not proving the answer is what it is, only eliminating the answers that can't be right. Regardless, this is not even remotely close to resembling a real GMAT question, even if a couple of the concepts in the solution might be useful on rare occasions.