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
Equal sums
This topic has expert replies
-
- Legendary Member
- Posts: 576
- Joined: Sat Mar 13, 2010 8:31 pm
- Thanked: 97 times
- Followed by:1 members
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.
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.
- outreach
- Legendary Member
- Posts: 748
- Joined: Sun Jan 31, 2010 7:54 am
- Thanked: 46 times
- Followed by:3 members
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
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
-------------------------------------
--------------------------------------
General blog
https://amarnaik.wordpress.com
MBA blog
https://amarrnaik.blocked/
--------------------------------------
General blog
https://amarnaik.wordpress.com
MBA blog
https://amarrnaik.blocked/
-
- Legendary Member
- Posts: 610
- Joined: Fri Jan 15, 2010 12:33 am
- Thanked: 47 times
- Followed by:2 members
x +ve integers - let the first digit be noutreach 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
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 ?
-
- Legendary Member
- Posts: 809
- Joined: Wed Mar 24, 2010 10:10 pm
- Thanked: 50 times
- Followed by:4 members
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.
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.
Can you please explain how you arrived at 10n+45 and 4m+6 ?kstv wrote:x +ve integers - let the first digit be noutreach 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
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 ?
Thanks!
- Treker
- Treker
-
- Legendary Member
- Posts: 809
- Joined: Wed Mar 24, 2010 10:10 pm
- Thanked: 50 times
- Followed by:4 members
you assume n and m are the first terms of respective seriestreker wrote:Can you please explain how you arrived at 10n+45 and 4m+6 ?kstv wrote:x +ve integers - let the first digit be noutreach 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
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 ?
-
- Senior | Next Rank: 100 Posts
- Posts: 44
- Joined: Sat Apr 10, 2010 8:33 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
Thanks
- ajith
- Legendary Member
- Posts: 1275
- Joined: Thu Sep 21, 2006 11:13 pm
- Location: Arabian Sea
- Thanked: 125 times
- Followed by:2 members
say n is the first numbernisha.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
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
Always borrow money from a pessimist, he doesn't expect to be paid back.