ITS 0.rattan123 wrote:What is the remainder when 2^100+3^100+4^100+5^100 is divided by 7
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remainder
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That's helpful to him I'm sure... if he's asking i'm sure he wants an approach to how to figure it out.killer1387 wrote:ITS 0.rattan123 wrote:What is the remainder when 2^100+3^100+4^100+5^100 is divided by 7
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Shalabh's Quants
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=> (2^100+3^100+4^100+5^100)/7rattan123 wrote:What is the remainder when 2^100+3^100+4^100+5^100 is divided by 7
=> (2.2^99 + 3.3^99 + 2^200 + 5.5^99)/7
=> [2*(2^3)^33 + 3*(3^3)^33 + 2^2*(2^3)^66 + 5*(5^3)^33]/7
=> [2*8^33 + 3*27^33 + 2^2*8^66 + 5*125^33]/7
=> [2*(7+1)^33 + 3*(28-1)^33 + 2^2*(7+1)^66 + 5*(126-1)^33]/7
=> Taking remainders, [2*1^33 + 3*(-1)^33 + 2^2*1^66 + 5*(-1)^33]/7
=> [2+ 3*(7-1)+ 4 + 5*(7-1)]/7; For [(-1)^n]/m; if n is odd, remainder is m - n.
=> [2 + 3*6 + 4 + 5*6]/7
=> 54/7 => Remainder is 5.
Shalabh Jain,
e-GMAT Instructor
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That is incorrect the answer is zero...Shalabh's Quants wrote:=> (2^100+3^100+4^100+5^100)/7rattan123 wrote:What is the remainder when 2^100+3^100+4^100+5^100 is divided by 7
=> (2.2^99 + 3.3^99 + 2^200 + 5.5^99)/7
=> [2*(2^3)^33 + 3*(3^3)^33 + 2^2*(2^3)^66 + 5*(5^3)^33]/7
=> [2*8^33 + 3*27^33 + 2^2*8^66 + 5*125^33]/7
=> [2*(7+1)^33 + 3*(28-1)^33 + 2^2*(7+1)^66 + 5*(126-1)^33]/7
=> Taking remainders, [2*1^33 + 3*(-1)^33 + 2^2*1^66 + 5*(-1)^33]/7
=> [2+ 3*(7-1)+ 4 + 5*(7-1)]/7; For [(-1)^n]/m; if n is odd, remainder is m - n.
=> [2 + 3*6 + 4 + 5*6]/7
=> 54/7 => Remainder is 5.
consider breaking down the question this way:-What is the remainder when 2^100+3^100+4^100+5^100 is divided by 7
Remainder of 2^100 / 7 = 2*(2^99) / 7
or 2*(2^3)^33 /7
or 2*8^33/7 = 2 (as remainder of 8*8*8..../ 7 = 1*1*1.... =1 only)
similarly R. of 3^100 / 7 or (3^2)^50 / 7 or 9^50 / 7 --- (1)
as remainder of 9 / 7 is 2 we can say R of (1) above = R of 2^50 / 7 or (2^2)*(2^3)^16 / 7
or 4*8^16 / 7 thus by same logic as above R. here is 4*1 = 4
Similarly, R. of 4^100 / 7 = 2^200 / 7 = 4*(2^3)^66 / 7 = 4*8^66 / 7 = 4
similarly, R. of 5^100 / 7 = R. of 25^50 / 7 = R. of 4^50 / 7 = R. of 2*(2^3)^33 = 2
adding all four together 2+4+4+2 = 12 thus finally remainder is 12 / 7 = 5
Mr. Anderson (:-D),Neo Anderson wrote:consider breaking down the question this way:-What is the remainder when 2^100+3^100+4^100+5^100 is divided by 7
Remainder of 2^100 / 7 = 2*(2^99) / 7
or 2*(2^3)^33 /7
or 2*8^33/7 = 2 (as remainder of 8*8*8..../ 7 = 1*1*1.... =1 only)
similarly R. of 3^100 / 7 or (3^2)^50 / 7 or 9^50 / 7 --- (1)
as remainder of 9 / 7 is 2 we can say R of (1) above = R of 2^50 / 7 or (2^2)*(2^3)^16 / 7
or 4*8^16 / 7 thus by same logic as above R. here is 4*1 = 4
Similarly, R. of 4^100 / 7 = 2^200 / 7 = 4*(2^3)^66 / 7 = 4*8^66 / 7 = 4
similarly, R. of 5^100 / 7 = R. of 25^50 / 7 = R. of 4^50 / 7 = R. of 2*(2^3)^33 = 2
adding all four together 2+4+4+2 = 12 thus finally remainder is 12 / 7 = 5
When I run this thru excel or a calculator there is no remainder... so what I think (which is prob 99% wrong) is you can factor out the power and get 2 + 3 + 4 + 5 which is 14; so 14/2 = 7 no remainder
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tomada
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(2^3)/7 -> Remainder = 1
(2^4)/7 -> Remainder = 2
(2^5)/7 -> Remainder = 4
(2^6)/7 -> Remainder = 1
(2^7)/7 -> Remainder = 2
(2^8)/7 -> Remainder = 4
...and this pattern continues...
(2^9)/7 -> Remainder = 1
(2^10)/7 -> Remainder = 2
When the exponent is 3,6,9,12, etc., the Remainder = 1
When the exponent is 4,7,10,13, etc., the Remainder = 2
When the exponent is 5,8,11,14, etc., the Remainder = 4
An exponent of 100 is part of the set that begins with 4,7,10,13 so, for (2^100)/7, the Remainder = 2
That's as far as I got.
(2^4)/7 -> Remainder = 2
(2^5)/7 -> Remainder = 4
(2^6)/7 -> Remainder = 1
(2^7)/7 -> Remainder = 2
(2^8)/7 -> Remainder = 4
...and this pattern continues...
(2^9)/7 -> Remainder = 1
(2^10)/7 -> Remainder = 2
When the exponent is 3,6,9,12, etc., the Remainder = 1
When the exponent is 4,7,10,13, etc., the Remainder = 2
When the exponent is 5,8,11,14, etc., the Remainder = 4
An exponent of 100 is part of the set that begins with 4,7,10,13 so, for (2^100)/7, the Remainder = 2
That's as far as I got.
I'm really old, but I'll never be too old to become more educated.
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Shalabh's Quants
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I am sure answer is 5 only. Hope you both too agree now.seal4913 wrote:That is incorrect the answer is zero...Shalabh's Quants wrote:=> (2^100+3^100+4^100+5^100)/7rattan123 wrote:What is the remainder when 2^100+3^100+4^100+5^100 is divided by 7
=> (2.2^99 + 3.3^99 + 2^200 + 5.5^99)/7
=> [2*(2^3)^33 + 3*(3^3)^33 + 2^2*(2^3)^66 + 5*(5^3)^33]/7
=> [2*8^33 + 3*27^33 + 2^2*8^66 + 5*125^33]/7
=> [2*(7+1)^33 + 3*(28-1)^33 + 2^2*(7+1)^66 + 5*(126-1)^33]/7
=> Taking remainders, [2*1^33 + 3*(-1)^33 + 2^2*1^66 + 5*(-1)^33]/7
=> [2+ 3*(7-1)+ 4 + 5*(7-1)]/7; For [(-1)^n]/m; if n is odd, remainder is m - n.
=> [2 + 3*6 + 4 + 5*6]/7
=> 54/7 => Remainder is 5.
Shalabh Jain,
e-GMAT Instructor
e-GMAT Instructor
The answer is 5 i made mistake while solving initially;
2^100+3^100+4^100+5^100
= 2^100+3^100+(7-3)^100+(7-2)^100
now taking remainders:
R{2*(2^100+3^100)}%7
= R{2(2+4)}%7
= 5
2^100+3^100+4^100+5^100
= 2^100+3^100+(7-3)^100+(7-2)^100
now taking remainders:
R{2*(2^100+3^100)}%7
= R{2(2+4)}%7
= 5
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ronnie1985
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2^100 = 2^99*2 = 8^33*2, R(2^100/7) = R(2*8^33/7) = 2
3^100 = 9^50, R(9^50/7) = R(2^50/7) = R(4*8^16/7) = 4
4^100 = 2^200, R(2^200/7) = R(4*8^66/7) = 4
R(5^100/7) = R((-2)^100/7) = R(2^100/7) = R(2*8^33/7) = 2
R(Expression/7) = R((2+4+4+2)/7) = 5
3^100 = 9^50, R(9^50/7) = R(2^50/7) = R(4*8^16/7) = 4
4^100 = 2^200, R(2^200/7) = R(4*8^66/7) = 4
R(5^100/7) = R((-2)^100/7) = R(2^100/7) = R(2*8^33/7) = 2
R(Expression/7) = R((2+4+4+2)/7) = 5
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the easiest way is taking 1^100 out as a common multiple we will get
1^100 (2+3+4+5)/7 = 1^100 (14)/7 the remainder is always zero...isnt it a quick way?
1^100 (2+3+4+5)/7 = 1^100 (14)/7 the remainder is always zero...isnt it a quick way?
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Hi Sana!sana.noor wrote:the easiest way is taking 1^100 out as a common multiple we will get
1^100 (2+3+4+5)/7 = 1^100 (14)/7 the remainder is always zero...isnt it a quick way?
I'm afraid powers don't work that way.
2^100 is NOT equal to 2*(1^100) (which is equal to 2).
So, we cannot take 1^100 common.
This is a very advanced level remainder problem which is unlikely to be tested in GMAT.
And the correct answer is 5 as shown by Shalabh/Neo/Ronnie.
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faraz_jeddah
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