Official Quant thread for CAT 2013

1> Remainder of 456456456..............900 digits when divided by 37 ?

2> 2^1096 when divided by 1096 then rem ?

@sachisurbhi said:
find the remainder when 84^79 is divided by 100a)24b)44c)64d)84
84^79 mod 100
100 = 25*4

84^79 mod 4 = 0

84^79 mod 25 = 9^79 mod 25
We know that, 9^80 mod 25 = 0

=> 9 mod 25 * 9^79 mod 25 = 1
9 mod 25 * R =1
=> R = 14

So, 84^79 mod 25 = 14

R = 4k = 25l + 14 = 64
@amresh_maverick said:
1> Remainder of 456456456..............900 digits when divided by 37 ?2> 2^1096 when divided by 1096 then rem ?
456456456 mod 37 = -1
10^3 mod 37 = 1 ..
There are 100 times such numbers so R = -100 mod 37 = 11

@amresh_maverick said:
1> Remainder of 456456456..............900 digits when divided by 37 ?2> 2^1096 when divided by 1096 then rem ?
2^1096 mod 1096 = 8* 2^1093 mod 137

2^1093 mod 137 = 2^5 mod 137 = 32

So, R = 8*32 = 256 ??
@ScareCrow28 said:
456456456 mod 37 = -110^3 mod 37 = 1 ..There are 100 times such numbers so R = -100 mod 37 = 11
@ScareCrow28 said:
2^1096 mod 1096 = 8* 2^1093 mod 1372^1093 mod 137 = 2^5 mod 137 = 32So, R = 8*32 = 256 ??
@amresh_maverick said:
1> Remainder of 456456456..............900 digits when divided by 37 ?2> 2^1096 when divided by 1096 then rem ?
OA :
1> 11
2>256
@mohitjain said:
there is a triangle ABC..dere is a point P inside the triangle such dat its distance from d three sides are 6,8 and 11..find the perimeter of the triangle??
Triangle equilateral hai kya? Else might not be unique answer I think?

regards
scrabbler

1> 2^469 + 3^268 when divided by 22 rem ?
2>1212121212......................... 300 digits when divided by 999 rem ?
3>How many 3 digit nos exist , the cubes of which end in 56 ?

@amresh_maverick said:

2> 2^1096 when divided by 1096 then rem ?

2^1096 mod 1096 ---> 1096 = 8*137
2^1096 mod 137 ----> E(137) = 136
2^136k mod 137 = 1
2^8 mod 137 = 256

2*(2^3)^365 mod 8 = 0

so R = 256 ?

@amresh_maverick said:
1> Remainder of 456456456..............900 digits when divided by 37 ?2> 2^1096 when divided by 1096 then rem ?
@ScareCrow28 dekh lo answer sahi hai nai

1= 11
2= 256
@amresh_maverick said:
1> 2^469 + 3^268 when divided by 22 rem ?
6 + 5 = 11?

regards
scrabbler

@amresh_maverick said:
1> 2^469 + 3^268 when divided by 22 rem ?2>1212121212......................... 300 digits when divided by 999 rem ?3>How many 3 digits nos exist , the cubes of which end in 56 ?
1)11?
3)36?
@amresh_maverick said:
1> 2^469 + 3^268 when divided by 22 rem ?2>1212121212......................... 300 digits when divided by 999 rem ?3>How many 3 digit nos exist , the cubes of which end in 56 ?
1 - 11
2 - 112 ??
@iLoveTorres said:
3)36?
Options for
How many 3 digits nos exist , the cubes of which end in 56 ?

9
18
36
27
@amresh_maverick said:
3>How many 3 digit nos exist , the cubes of which end in 56 ?
18? All ending in 36 and 86?

regards
scrabbler

@amresh_maverick said:
1> 2^469 + 3^268 when divided by 22 rem ?

2^469 mod 22 = 2^469 mod 2*11
2^469 mod 11 = 2^9 mod 11 = 6

3^268 mod 2 = 1
3^268 mod 11 = 3^8 mod 11 = 81*81 mod 11 = 4*4 mod 11 = 16 mod 11 = 5

R = 6 + 5 mod 22 = 11 ?

@amresh_maverick said:
1> Remainder of 456456456..............900 digits when divided by 37 ?2> 2^1096 when divided by 1096 then rem ?
1)0
2)256
@amresh_maverick said:

3>How many 3 digit nos exist , the cubes of which end in 56 ?

18 ?

no. ending with 36 and 86

@sachisurbhi said:
find the remainder when 84^79 is divided by 100a)24b)44c)64d)84
@techgeek2050 said:
2 - 112 ??
Don't think it is 112 as 121212....300 digits is divisible by 9 and so is 999, so the remainder also should be a multiple of 9
@amresh_maverick said:
2>1212121212......................... 300 digits when divided by 999 rem ?

Sirji options kya they original mein?

regards
scrabbler

@scrabbler said:
Don't think it is 112 as 121212....300 digits is divisible by 9 and so is 999, so the remainder also should be a multiple of 9Sirji options kya they original mein?regardsscrabbler
@amresh_maverick i think it is 108