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25 May 2022, 01:30
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C
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Difficulty:
95%
(hard)
Question Stats:
31% (02:36) correct
69%
(02:09)
wrong
based on 171
sessions
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Let N = 123456789101112...4344 be the 79-digit number that is formed by writing the integers from 1 to 44 in order, one after the other. What is the remainder when N is divided by 45?
A. 1
B. 4
C. 9
D. 18
E. 44
Are You Up For the Challenge: 700 Level Questions
A. 1
B. 4
C. 9
D. 18
E. 44
Are You Up For the Challenge: 700 Level Questions
21 Jun 2022, 21:52
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Bunuel
rdrdrd1201
Fun one, but (probably) tougher than GMAC would use unless you're at 50 or 51 Quant...or unless I'm missing some trick (aside from mods other than 10 because GMAC doesn't test mods)!!!
In order to be divisible by 45, a number must be divisible by both 5 and 9. We can tell right away that our number is not divisible by 5. Let's see if it is divisible by 9.
Divisibility by 9 occurs when the sum of the digits is divisible by 9.
Let's look at the units digits in 1 through 44 first. The units digits from any ten consecutive numbers is divisible by 9 (each ten consecutive integers includes the following units digits: 0+9, 1+8, 2+7, 3+6, 4+5). So that takes care of the units digits from 1 through 40. Then we have 1+2+3+4 = 10 as the units digits from 41-44.
How about the tens digits? We have ten 1s, ten 2s, ten 3s, and five 4s. 10(1) + 10(2) + 10(3) + 5(4) = 80.
10 from the units digits + 80 from the tens digits = 90. That's divisible by 9, too, so our 79-digit number is divisible by 9!!
But our 79-digit number is not divisible by 5.
What if we subtract 9 from our 79-digit number? It would still be divisible by 9. And the units digit would now be 5, so that new number would be divisible by 5, too. That means that the new number is divisible by both 5 and 9, so the new number is a multiple of 45. Therefore, our 79-digit number is greater by 9 than a multiple of 45. So, the remainder of the 79-digit number when divided by 45 will be 9.Answer choice C.
General Discussion
10 Sep 2022, 09:23
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I won't discuss whether this type of question will be tested on the GMAT or not; I though believe such concept won't be tested.
We will break 45 into two co prime factors 9 and 5.
With 5 as divisor, we get 4 as the remainder as the last digit of the 79 digit number is 4.
To find remainder when the number is divided by 9, we can simply add the numbers from 1 to 44.
1+2+3+.....+44=(44*45)/2=22*45. This is clearly divisible by 9 giving 0 as the remainder.
With 5 remainder is 4 or 9 or 14 or 19 etc.
With 9 remainder is 0 or 9 or 18 or 27 etc.
Therefore, Remainder of the no when divide by 45 will be 9
We will break 45 into two co prime factors 9 and 5.
With 5 as divisor, we get 4 as the remainder as the last digit of the 79 digit number is 4.
To find remainder when the number is divided by 9, we can simply add the numbers from 1 to 44.
1+2+3+.....+44=(44*45)/2=22*45. This is clearly divisible by 9 giving 0 as the remainder.
With 5 remainder is 4 or 9 or 14 or 19 etc.
With 9 remainder is 0 or 9 or 18 or 27 etc.
Therefore, Remainder of the no when divide by 45 will be 9
