CAT Questions on Last Two Digits

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CAT Questions on Last Two Digits
CAT Questions on Last Two Digits

CAT Questions on Last Two Digits:

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Question 1: If the last two digits of (1!+2!+3!….. 50!) are ab, then find the value of a*b?

a) 3
b) 4
c) 5
d) 6

Question 2: Find the last two non-zero digits of 25!.

a) 74
b) 94
c) 84
d) 64

Question 3: Find the sum of the last two digits in $89^{82}$.

a) 1
b) 3
c) 5
d) 7

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Question 4: What are the last two digits of the sum $(19!)^2 + (18!)^2 + (17!)^2 + ………(0!)^2$

a) 33
b) 34
c) 17
d) 18

Question 5: Find the last two digits of the number $27^{103}$?

a) 63
b) 73
c) 93
d) 83

Solutions for CAT Questions on Last Two Digits:

Solutions:

1) Answer (A)

Lat two digits is nothing but the remainder with 100
From 10! onwards all terms will be divisible by 100
So we just need to find out the remainder of (1! + 2! + ….9! ) with 100
So on calculating the remainders we get,
1+2+6+24+20+20+40+20+80 = 213,
213mod 100 = 13, so last two digits are 13
Hence the product of digits is 1*3 = 3

2) Answer (C)

The last two non-zero digits of the factorial of a number X, which is divisible by 5 can be calculated as below.
last two non-zero digits of X! = last two non zero digits of $[x/5]! * 12^{x/5}$
Here x = 25. ==> $[25/5]! *12^{25/5}$ = $5!*12^{5}$ = 120*32 (last two digits of $12^5$ is 32) ==> 84
Therefore last two digits of 25! = 84.

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3) Answer (B)

Last two digits can be found by finding the remainder with 100.
$89^{82}=(90-1)^{82}$
$=(…00)-(82\times90)+1$
$=(…00)-7380+1$
$=(…00)-7379$
$=(…21)$
So the last two digits are 21.
Sum of digits = 3

4) Answer (D)

From $(5!)^2$ to $(19!)^2$ every number in the series ends with ‘00’.
Now we need to calculate the sum of last two digits of numbers from $(0!)^2$ to $(4!)^2$ = 1 + 1 + 4 + 36 + 576 = 618
Therefore last two digits of the sequence are 18.

5) Answer (D)

Last two digits of a number is nothing but the remainder when the number is divided by 100. So we essentially have to find the remainder when the given number is divided by 100.
$27^{103}$ mod 100
100 can be written as 25*4
$27^{103}$ mod 4 = $(-1)^{103}mod4 = -1 = 3$
$27^{103}$mod 25 = $2^{103}$mod25 = $(2^{10})^{10}*8$mod25
= $(1024)^{10}*8$mod25 = 1*8 = 8
Thus the required remainder is of the form
4k+3 = 25m+8
If m = 1, we get 33 but this is not of the form 4k+3. Hence this is not possible.
If m = 2, we get 58 but this is again not of the form 4k+3. Hence this is also not possible.
m=3 gives 83. This is of the form 4k+3. Hence the required remainder is 83.

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