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If 22^3+23^3+24^3+....

Expert replies
Source: — Problem Solving |

by Brent@GMATPrepNow » Wed Oct 24, 2012 7:36 am
nasheen wrote:If 22^3+23^3+24^3+....+87^3+88^3 is divided by 110 then the remainder will be

A. 55
B.1
C.33
D.0
E.44

A
Proviso: Unless there's a much easier solution out there, I believe that this question is out of scope for the GMAT.

First notice that there are 57 terms in the given expression (since 88-22+1 = 57)
Now notice that we can rewrite the sum as:
(22^3 + 88^3) + (23^3 + 87^3) + (24^3 + 88^3) . . . + (54^3 + 56^3) + 55^3
Why did I do this?

Well, there's a nice factoring technique called sum of cubes factoring. It goes like this:
a^3 + b^3 = (a + b)(a^2 - ab + b^2)

IMPORTANT: You do not need to know the technique of sum of cubes factoring for the GMAT. For this reason, I believe this question is out of scope (unless there's a much easier solution that I've overlooked).

Moving along, we can apply this factoring technique to each of the sum of cubes in our rearranged expression. For example:
22^3 + 88^3 = (22 + 88)(22^2 - (22)(88) + 88^2)
= (110)(some other value)
Notice that our sum of cubes can be rewritten as 110 times some other number.
This means that (22^3 + 88^3) is divisible by 110, which means the remainder will be 0 when we divide (22^3 + 88^3) by 110

Using similar logic, we can show that (23^3 + 87^3) is divisible by 110 (since 23+87=110, which means 110 will be part of the factorization . . . try it)
And we can show that (24^3 + 86^3) is divisible by 110.
And so on.

So, almost all of the terms in (22^3 + 88^3) + (23^3 + 87^3) + (24^3 + 88^3) . . . + (54^3 + 56^3) + 55^3 are divisible by 110.

So, if we were to ignore the 55^3, the sum ((22^3 + 88^3) + (23^3 + 87^3) + (24^3 + 88^3) . . . + (54^3 + 56^3) would have remainder 0 when divided by 110

So, all we need to do now is determine the remainder when 55^3 is divided by 110
Even this step is a bit of work since it requires some modular arithmetic that we don't need for the GMAT.

Alternatively, I suppose one could evaluate 55^3 to get 166,375 and then divide by 110 to get a remainder of 55 (A)

BUT, as I said, this question is out of score for the GMAT (unless I've overlooked a much easier solution).

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by GMATGuruNY » Wed Oct 24, 2012 9:04 am
nasheen wrote:If 22^3+23^3+24^3+....+87^3+88^3 is divided by 110 then the remainder will be

A. 55
B.1
C.33
D.0
E.44

A
Brent's solution is great. Here's an alternate approach:

When positive integer x is divided by 110 -- or by ANY multiple of 10 -- the units digit of the remainder will always be equal to the units digit of x.
If x=223, 223/110 = 1 R 113.
If x=587, 587/120 = 4 R 107.
If x=955, 955/130 = 7 R 45.
In each case, the units digit of the remainder is equal to the units digit of x.

Since each of the answer choices here offers a different units digit, we can determine the correct answer simply by calculating the units digit of 22³+...+88³.

When the digits 2 through 1 are cubed, we get the following cycle of units digits:
2³ --> 8.
3³ --> 7.
4³ --> 4.
5³ --> 5.
6³ --> 6.
7³ --> 3.
8³ --> 2.
9³ --> 9.
0³ --> 0.
1³ --> 1.

The sum of the units digits of the cycle = 8+7+4+5+6+3+2+9+0+1 = 45.
From 22³ to 71³, this cycle will repeat 6 times:
6*45 = 250, implying that 22³+...+71³ will have a units digit of 0.

The sum of the units digits of 72³+...+88³ = 8+7+4+5+6+3+2 = 35, implying that the units digit of this portion of the sum will be 5.

Thus, the sum of the units digits for the ENTIRE sum = 0+5 = 5.
Since the units digit of the sum is 5, when the sum is divided by 110, the units digit of the remainder will also be 5.

The correct answer is A.
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by Brent@GMATPrepNow » Wed Oct 24, 2012 9:09 am
Great solution (as always), Mitch!
I tried to solve it without considering the answer choices. Fortunately, only one answer choice has units digit 5. Otherwise, the question would be a total pain.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by smanstar » Sat Oct 27, 2012 4:27 am
a^n + b^n is divisible by (a+b) is n is odd.

Since in this question we find 22 + 88 = 110 , thus all the terms will be divisible by 110, except for the middle term ( as 22- 88 we have odd no of numbers) which is 55^3

So remainder is (55*55*55)/ 110 which is 55 itself.
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