Let N = 123456789101112...4344 be the 79-digit number that is formed

Question

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

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ThatDudeKnows · Accepted Answer

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.

rdrdrd1201 · Answer

i reckon the end of 
- 45 x odd ends with 5 
- 45 x even ends with 0 
Thus the remainder will either be 
4-5 = ends with 9 OR (Option C)
4-0 = ends with 4 (Option B/E) but i cant find a pattern to determine this further - can someone help explain this please! thanks a lot

rdrdrd1201 · Answer

i reckon the end of 
- 45 x odd ends with 5 
- 45 x even ends with 0 
Thus the remainder will either be 
4-5 = ends with 9 OR (Option C)
4-0 = ends with 4 (Option B/E) but i cant find a pattern to determine this further - can someone help explain this please! thanks a lot

ABHISEKGMAT123 · Answer

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

HarshavardhanR · Answer

My two cents (How I solved this). This is a super-fun question!

N = 1234....4344 (number formed by writing integers from 1 to 44 in order).

To find:  Remainder when N is divided by 45.

For a number to be divisible by 45, it should be divisible by both 5 and 9.

Point:  If we can find the closest number to N which is divisible by 5 (ends with 5 or 0) and also divisible by 9 (sum of all digits = 9), from there, we can easily find the remainder when N is divided by 45.

(1) Let's check if N is divisible by 9.

Sum of digits =  sum of unit digits  +  sum of tens digits

Sum of units digits  = Sum of units digits from 1 to 9 + sum of units digits from 10 to 19 + sum of units digits from 20 to 29 +.... sum of units digits from 40 to 44.

=   +   +   +   +

Sum of digits from 0 to 9 = 45.

Sum of unit digits  = 4x45 + 10 = 180+10 = 190.

Sum of tens digits = sum of tens digits of 10 to 19 + sum of tens digits of 20 to 29 + sum of tens digits of 30 to 39 + sum of tens digits of 40 to 44

Sum of tens digits  = 10(1) + 10(2) + 10(3) + 5(4) = 80.

Therefore,
Sum of digits of N = 190 + 80 = 270 = multiple of 9.

Therefore,  N is divisible by 9.

Now, close to N, if we can find a number that is divisible by 9 but also by 5 (ending with 5 or 0), then, that number is divisible by 45.

Observe:

1234....4335 , 1234........4344, 1234......4353, 1234.........4362, 1234.......4371,  1234..........80 .

We can use either of the two numbers highlighted above.

1234......4335 is divisible by 45 because it is divisible by both 9 and 5. Our N is 9 more than this number. Hence, the remainder when N is divided by 45 should be 9.

Alternately, 1234....4380 is divisible by 45. Our N is 36 less than this number. Hence, the remainder when N is divided by 45 should be -36 i.e., +(45-36) = 9.

Choice C.

Hope this helps!
Harsha

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Let N = 123456789101112...4344 be the 79-digit number that is formed

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Let N = 123456789101112...4344 be the 79-digit number that is formed

Prep Toolkit

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