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Equal sums

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by neoreaves » Wed Apr 07, 2010 5:02 am
If a is the sum of x consecutive positive integers. b is the sum of y
consecutive positive integers. For which of the following values
of x and y is it impossible that a = b?

a) x = 2; y = 6
b) x = 3; y = 6
c) x = 7; y = 9
d) x = 10; y = 4
e) x = 10; y = 7
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by liferocks » Wed Apr 07, 2010 7:31 am
hi,

I have used following method,
let first term of a be m and b be n
so if a=b we will get
x/y=(2n+(y-1))/(2m+(x-1))

for all option other than d there is valid integer value of m and n present
hence ans .d
can you please confirm the method and OA.
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by outreach » Wed Apr 07, 2010 9:33 am
sum of A will not equal sum B if one group can have an even sum and the other can have an odd sum. The only choice that meets that requirement is D.

In D sum of every 10 consecutive integers ends with 5 (as the unit digit). and sum of every 4 consecutive integers ends with an even digit
neoreaves wrote:If a is the sum of x consecutive positive integers. b is the sum of y
consecutive positive integers. For which of the following values
of x and y is it impossible that a = b?

a) x = 2; y = 6
b) x = 3; y = 6
c) x = 7; y = 9
d) x = 10; y = 4
e) x = 10; y = 7
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by RJ43 » Wed Apr 07, 2010 12:12 pm
EDIT misread "impossible" as "possible"
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by kstv » Wed Apr 07, 2010 9:44 pm
outreach wrote:In D sum of every 10 consecutive integers ends with 5 (as the unit digit). and sum of every 4 consecutive integers ends with an even digit
x +ve integers - let the first digit be n
y + ve integers - lets the first digit be m
Option D d) x = 10; y = 4
x integers - if 1st digit is n the sum of the series is 10n +45 has to be a odd number for all values of n

y integers - if 1st digit is m the sum of the series is 4m+6 has to be a even number for all values of m
Great solution but still we have to check each option. Is it possible to hone on to the correct choice in 2 mins ?
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by neoreaves » Wed Apr 07, 2010 10:18 pm
OA: D
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by akhpad » Thu Apr 08, 2010 3:13 am
If a is the sum of x consecutive positive integers. b is the sum of y consecutive positive integers. For which of the following values
of x and y is it impossible that a = b?

I will just do for option D

Lets assume first term be A and B for respective series.
a = 10A + 45 = Odd Number
b = 4B + 6 = Even Number

a = b Cannot be

10A + 39 = 4B

We cannot get a number, which has unit digit 9 and multiple of 4

Hence a=b impossible.
Last edited by akhpad on Thu Apr 08, 2010 3:28 am, edited 1 time in total.
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by treker » Thu Apr 08, 2010 3:21 am
kstv wrote:
outreach wrote:In D sum of every 10 consecutive integers ends with 5 (as the unit digit). and sum of every 4 consecutive integers ends with an even digit
x +ve integers - let the first digit be n
y + ve integers - lets the first digit be m
Option D d) x = 10; y = 4
x integers - if 1st digit is n the sum of the series is 10n +45 has to be a odd number for all values of n

y integers - if 1st digit is m the sum of the series is 4m+6 has to be a even number for all values of m
Great solution but still we have to check each option. Is it possible to hone on to the correct choice in 2 mins ?
Can you please explain how you arrived at 10n+45 and 4m+6 ?
Thanks!
- Treker
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by akhpad » Thu Apr 08, 2010 3:29 am
treker wrote:
kstv wrote:
outreach wrote:In D sum of every 10 consecutive integers ends with 5 (as the unit digit). and sum of every 4 consecutive integers ends with an even digit
x +ve integers - let the first digit be n
y + ve integers - lets the first digit be m
Option D d) x = 10; y = 4
x integers - if 1st digit is n the sum of the series is 10n +45 has to be a odd number for all values of n

y integers - if 1st digit is m the sum of the series is 4m+6 has to be a even number for all values of m
Great solution but still we have to check each option. Is it possible to hone on to the correct choice in 2 mins ?
Can you please explain how you arrived at 10n+45 and 4m+6 ?
you assume n and m are the first terms of respective series
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by nisha.menon294 » Wed Apr 14, 2010 2:02 am
after assuming n and m to be the first number of the series , what is the formula you have used to arrive at 10 n + 45 ?

Thanks
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by ajith » Wed Apr 14, 2010 4:36 am
nisha.menon294 wrote:after assuming n and m to be the first number of the series , what is the formula you have used to arrive at 10 n + 45 ?

Thanks
say n is the first number
the last number would be n+9

The sum of the Arithmetic Progression with 'a' as the first term 'b' as the last term and with 'p' terms is p/2 (a +b)

Applying the same formula here Sum of terms = 10/2 (n+ n+9) = 5(2n +9) = 10n +45
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