ANSWERS
1. C
2. D
3. C
SOLUTIONS
1.
For n to be divisible by 10, y has to be zero. So, n = 4x130
As suggested by Karishma above, when doing the divisibility test for 3, we don't need to add all digits: 4+x+1+3+0. We can simply add the digits that are non-multiples of 3 or whose sum is not a multiple of 3. So, in this case, we need to only add 4+x+1 = 5+x
For what values of x will 5+x be divisible by 3? For x=1, 4, 7.
So, in all there are only 3 possible numbers that will be divisible by both 3 and 10:
41130, 44130, 47130
(C) is the answer.
2.
For n to be divisible by 5, y could either be 0 or 5.
Case 1: y= 0So, n= 4x130
Using the divisibility rule for 3, we find that possible values of x are 1, 4, 7. However, it is given that x>1. So, in total, only 2 values of x are possible.
Case 2: y=5So, n= 4x135
Here sum of 1+3+5 = 9, a multiple of 3. So, we need to only consider the digits 4 and x for the divisibility test.
Possible values of x are 2, 5, 8. So, 3 values are possible.
Thus, total possible values of n are 5:
44130, 47130
42135, 45135, 48135
3.
For n to be divisible by 2, y= 0, 2,4,6,8. However, it is given that y>2. So, allowed values of y: 4, 6, 8
Case 1: y= 4So, n= 4x134
Possible values of x = 0,3,6,9. Therefore, 4 values.
Case 2: y=6So, n= 4x136
We need to only consider the digits 4, x, 1 for the divisibility rule (since 3 and 6 are already multiples of 3).
Possible values of x= 1,4,7. Therefore, 3 values
Case 3: y=8So, n= 4x138
We need to consider only the digits 4 and x for the divisibility rule (since 3 and 8+1 are already multiples of 3).
Possible values of x = 2, 5, 8. Therefore, 3 values
Total values of n possible = 4+3+3 = 10