Number System

Given: 6735Q1
6 + 7 + 3 + 5 + Q + 1 = 22 + Q
The least value of Q when 22 + Q divisible by 9.
Let, the least value of Q is 5.

Let the three digit number, xyz = 100x + 10y + z
After second and third digit interchanged, xzy = 100x + 10z + y
So, xyz - xzy = 45
(100x + 10y + z) - (100x + 10z + y) = 45
100x + 10y + z - 100x - 10z - y = 45
9y - 9z = 45
y - z = 5
∴ difference between last two digit of that number is 5.

We know, Sum of first 130 numbers= 1 + 2 + 3 + ....... + 130
1 + 2 + 3 + ......+ n = n(n + 1)/2
Here, n = 130
The sum of first 130 numbers = 130(130 + 1)/2
= 130(131)/2
= 8515

When we devided 91 by 13 remainder is zero.
So, 17/13 remainder is 4.

we know that X is twice of Y.
(3X + 4Y)/2 > 50
3X + 4Y > 100
10x > 100
Y > 10
Multiplying 2 on both sides,
2Y > 20
X > 20 (Since, 2Y = X )
∴ the smallest possible integer value of X is 21.

Let the numbers be×and y.
3x + 2y = 20 ……(i)
2x + 3y = 15 ……(ii)
After solving, we get
x = 6 ; y = 1
So, the largest number is 1.

Details

Unit digit in (6374)¹⁷⁹³×(625)³¹⁷×(341)⁴⁹¹
= Unit digit in (4)¹⁷⁹³×(5)³¹⁷×(1)⁴⁹¹
= Unit digit in 4×5×1
= Unit digit is 0

Numbers are decreases by 24
The Next Number should decrease by 24 from the previous number.
So, 101- 24 = 77

Let the three consecutive odd numbers be x,×+ 2,×+ 4.
We know that,
x = x + 2 = x + 4 = x + 20
x = 7
Hence, three odd numbers are 7, 9, 13 and the middle number is 9.