Combinatorics DPP - RACE 8 - pending from Q3 onwards

Q1. How many positive integers 'n' are there such that 'n' is a divisor of one of the numbers 10^40, 20^30?
Solution:
Factorize 10^40 = 2^40.5^40
Number of divisors = 41.41 = 1681
20^30 = 2^60.5^30
Number of divisors = 61.31 = 1891

What is their gcd?
2^40.5^30
Number of divisors of gcd = 41*31

Total divisors of the given numbers  = 1681 + 1891
But the divisors of gcd would be counted twice so subtract it: 1681 + 1891 - 41*31 = 2301



Q2. Consider all 6-digit numbers of the form abccba where b is odd. Determine the number of all such 6-digit numbers that are divisible by 7.
Before solving this, try solving this similar question:

Q. How many 3 digit numbers like aba are divisible by 3?
Solution:

100a+10b+a = 0 mod 3
101a + 10b = 0
2a + b = 0
-a + b = 0 => difference of a and b is divisible by 3.
=> remainder of a and b div 3 is same.

a != 0
a = 1,4,7 b = 1,4,7
a = 2,5,8 b = 2,5,8
a = 3,6,9 b = 0,3,6,9

Total 9 + 9 + 12 = 30

Solution to original question:

abccba = (100,001)a + (10,010)b + 1100c
If this is divisible by 7, then:
Div mod 7 =>
0 = 6a + 0b + 1c
0 = -1.a + 1.c => c - a = 0(div 7)
Make pairs:
a can't be 0
c-a = 7
(a,c) = (1,8) (2,9)
a-c = 7
(a,c) = (9,2) (8,1)
a = c
(1,1)(2,2)....(9,9)
Total 14 pairs.
b can have 5 values.
So total 70 is the answer.

Q3: Let E denote the set of all natural numbers n such that 3 < n < 100 and the set {1, 2, 3, ......n} can be partitioned into 3 subsets with equal sums. Find the number of elements of E.
Solution:

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