Five questions. Five minutes. Your time starts now!
Q. 1 What are the last two digits of 7^2008? (CAT 2008)
(1) 21 (2) 61 (3) 01 (4) 41 (5) 81
Q.2 A new flag is to be designed with six vertical stripes using some or all of the colors yellow, green, blue, and red. Then the number of ways this can be done so that no two adjacent stripes have the same color is (CAT 2004)
(1) 12 × 81 (2) 16 × 192 (3) 20 × 125 (4) 24 × 216
Q.3 Consider the set S = {2, 3, 4… 2n+1}, where n is a positive integer larger than 2007. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of X – Y? (CAT 2007)
(1) 0 (2) 1 (3) n/2 (4) (n+1)/2n (5) 2008
Q.4 Let S be the set of prime numbers greater than or equal to 2 and less than 100. Multiply all elements of S. With how many consecutive zeroes will the product end? (CAT 2000)
(1) 1 (2) 4 (3) 5 (4) 10
Q.5 What is the sum of all two-digit numbers that give a remainder of 3 when divided by 7? (CAT 2003)
(1) 666 (2) 676 (3) 683 (4) 777
Solutions:
Q. 1 What are the last two digits of 7^2008? (CAT 2008)
Solution:
7 follows the cyclicity of 4 when it comes to units places of powers. 7^1 = 7, 7^2 = 9, 7^3 = 3, 7^4 = 1
Interestingly, 7 also follows cyclicity for tens place and units place. 7^1 = 07, 7^2 = 49, 7^3 = 343, 7^4 = 2401
As 2008 is a multiple of 4, the answer is 01.
Q.2 A new flag is to be designed with six vertical stripes using some or all of the colors yellow, green, blue, and red. Then the number of ways this can be done so that no two adjacent stripes have the same color is (CAT 2004)
Solution:
The best way is to simply draw the flag. The first can be colored using any of the 4 colors. For each of the rest of the stripes, we will be left with 3 options.
4 | 3 | 3 | 3 | 3 | 3
Answer = 4 × 3^5 = 12 × 81
Q.3 Consider the set S = {2, 3, 4… 2n+1}, where n is a positive integer larger than 2007. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of X – Y? (CAT 2007)
Solution:
S = {2, 3, 4… 2n+1} = {even, odd, even… odd}
Assuming a hypothetical case. Let say S = {2, 3, 4, 5}
In this case, X = (3+5)/2 = 4 and Y = (2+4)/2 = 3 → X – Y = 1. This will be true for S = {2, 3, 4… 2n+1}
Answer = 1
Q.4 Let S be the set of prime numbers greater than or equal to 2 and less than 100. Multiply all elements of S. With how many consecutive zeroes will the product end? (CAT 2000)
Solution:
To create a zero, one needs one power of 2 and one power of 5. Set of S = {2, 3, 5, 7, 11… 97}
In this set, we only have one 2 and one 5 giving us a zero at the units place. As rest of the numbers will not be divisible by 2 or 5, we will not have additional zeroes. Or, one can start multiplying from 2 × 3 = 6. 6 × 5 = 30. 30 × 7 = 210. 210 × 11 = 2310. With just by observation, we can figure out that the product is going to end with a single zero. Answer = 1
Q.5 What is the sum of all two-digit numbers that give a remainder of 3 when divided by 7? (CAT 2003)
Solution:
There are multiple approaches to solve this. First, is to actually add all the numbers. The first number to satisfy the condition is 10, and the last is 94. The second approach is to use the formula of Sum of n terms of an A.P.
Had the two-digit condition not been there, the number of terms would have been 14. Because of the condition, number of terms are 13.
Answer = 676
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