mgmat practice question

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varunkh70: Senior | Next Rank: 100 Posts; Posts: 51; Joined: Thu May 01, 2008 7:26 am; Thanked: 1 times

by varunkh70 » Wed Jul 08, 2009 3:17 pm

Is the sum of integers from 54 to 153,inclusive, divisible by 100?

Approach 1(explanation in mgmat guide) -

No. of terms =>153 - 54+1=100. No. of terms are even so special sums rule does not hold therefore the answer is "NO"

Special sums rule - the sum of odd no. of integers is always divisible by the no. of terms in the set but this does NOT work for even no. of terms(which is our case)

Approach 2(trying out the other way coz of which I get a different answer)
1) No. of terms is 100
2) Average of the 1st and last term is (54+153)/2= 103.5
3) 100 * 103.5 gives a multiple of 100
therefore the answer is "YES"

Can anyone explain the difference in the results?

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Source: Beat The GMAT — Quantitative Reasoning | Wed Jul 08, 2009 3:17 pm

ssmiles08: Master | Next Rank: 500 Posts; Posts: 472; Joined: Sun Mar 29, 2009 6:54 pm; Thanked: 56 times

Re: mgmat practice question

by ssmiles08 » Wed Jul 08, 2009 3:45 pm

varunkh70 wrote:Is the sum of integers from 54 to 153,inclusive, divisible by 100?

Approach 1(explanation in mgmat guide) -

No. of terms =>153 - 54+1=100. No. of terms are even so special sums rule does not hold therefore the answer is "NO"

Special sums rule - the sum of odd no. of integers is always divisible by the no. of terms in the set but this does NOT work for even no. of terms(which is our case)

Approach 2(trying out the other way coz of which I get a different answer)
1) No. of terms is 100
2) Average of the 1st and last term is (54+153)/2= 103.5
3) 100 * 103.5 gives a multiple of 100
therefore the answer is "YES"

Can anyone explain the difference in the results?

I think your first approach is right.

For your second approach: the "SUM" is 10350

10350 is NOT divisible by 100. for a number to be divisible by 100, it must have at least 2 zeros at the end and the result should be an integer.

Last edited by ssmiles08 on Wed Jul 08, 2009 3:47 pm, edited 1 time in total.

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varunkh70: Senior | Next Rank: 100 Posts; Posts: 51; Joined: Thu May 01, 2008 7:26 am; Thanked: 1 times

by varunkh70 » Wed Jul 08, 2009 3:46 pm

oh..my bad i got it now..10350 is just not divisible by 100..

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sreak1089: Master | Next Rank: 500 Posts; Posts: 379; Joined: Wed Jun 03, 2009 3:05 amFollowed by: 1 member; Thanked: 19 times; GMAT Score:690

by sreak1089 » Fri Jul 10, 2009 6:10 am

I do it this way :

54 + 55 + .. + 153 ==

1 + 2 + 3 + .. *+ 153 - (1 + 2 + 3 + .. .+ 53)

= (153 * 154) / 2 - (53 * 54) / 2
= 1/2 [ (100 + 53)(100 + 54) - 53 * 54) ]
= 1/2 (100*100 + 100 * 54 + 100 * 53)

when divided by 100:
= 1/2(100 + 54 + 53)

hence not divisibly by 100

Is there a shorter method?

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GMAT/MBA Expert

Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members; Location: Montreal; Thanked: 1090 times; GMAT Score:780

by Ian Stewart » Fri Jul 10, 2009 8:47 am

There are many ways to add evenly spaced sums; if you have a method that works for you and you're comfortable with it, there's no reason to learn a new one. They can all be done without any formulas as follows:

54 + 55 + 56 + ... + 151 + 152 + 153
= (53 + 1) + (53 + 2) + (53 + 3) + ... + (53 + 98) + (53 + 99) + (53 + 100)
= 53n + (1 + 2 + 3 + ... + 98 + 99 + 100)

where n is the number of terms. From the sum in brackets, we can see that n = 100, so our sum is

5300 + 101*50 = 5300 + 5050 = 10350.

(I added 1 + 2 + 3 + ... + 98 + 99 + 100 by regrouping as follows: (1 + 100) + (2 + 99) + (3 + 98) + ... = 101 + 101 + 101 + ... + 101 = 50*101, because we have 50 pairs of numbers which each add to 101).

Of course there are formulas you can learn for these problems that also work well.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com

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sreak1089: Master | Next Rank: 500 Posts; Posts: 379; Joined: Wed Jun 03, 2009 3:05 amFollowed by: 1 member; Thanked: 19 times; GMAT Score:690

by sreak1089 » Fri Jul 10, 2009 8:56 am

Hi Ian,

That was cool....I really appreciate your logic.

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