Theory of Numbers Test - 6

Single digit numbers are: 2, 4, 6, 8
Two digit numbers are: 4 × 5 = 20

(Because the first numeral can be any one from 2, 4, 6, 8 and the second numeral can be any
one from 0, 2, 4, 6, 8.

So, total 4 × 5 = 20 numbers). Required answer = 4 + 20 = 24.

If the two-digit number is ab, then a + b > ab

This is possible, only if

case 1: b is 0 and a ∈ [1, 9], such numbers are 10, 20……….., 90.
case 2 : a is 1 and b ∈ [1, 9], such numbers are 11, 12………., 19.
case 3 : b is 1 and a ∈ [2, 9], such numbers are 21, 31, 41…….., 91.

So, such numbers are 10, 20……….., 90 and 11, 12………., 19 and 21, 31, 41…….., 91.

So, total numbers are 9 + 9 + 8 = 26.

Let the 4-digit number n² be of the form aabb.
Now, we find that the sum of digits at even places = sum of digits at odd places = a + b each

Thus, the number is divisible by 11.
This implies that 11 is a factor of n²

Thus, 11 is a lso a factor of 'n'.
Now, 100² = 10000 which is the smallest 5-digit number

Also, 31² = 961, and 32² = 1024
Thus, n > 32 and n = 11b

Trying out diferent values of 'n' by putting b = 3, 4, 5, 6, 7, 8 and 9, we find that n² is of the form aabb only when b = 8 which gives us 'n' as 88 and n² as 7744

Thus, there is only one such 4-digit number.

From the options, 5.51 × 72 = 396.72

1944 = 2³ × 3⁵
So, to make it a perfect square, we have to multiply it with 2 × 3 = 6.

Since n is not a multiple of 7, it can take the following six forms:
(7k + 1), (7k + 2), (7k + 3), (7k + 4), (7k + 5) and (7k + 6)

But (7k + 6) will give a remainder 1, when divided by 7 and it is raised to even powers. Thus, m is even.

Thus, m can be either 2, 4 or 6.
Now, (7k + 2) raised to 2 when divided by 7 becomes 2²/7 and gives a remainder 4, not 1.

Also, (7k + 2) raised to 4 when divided by 7 becomes 2⁴/7 and gives a remainder 2, not 1.

But, (7k + 2) raised to 6 when divided by 7 becomes 2⁶/7 and gives a remainder 1.
Hence, m = 6

The first light blinks after 20 seconds or 3 times in a minute.
The second light blinks 5 times in two minutes or after a time period of 24 seconds.
LCM of (20, 24) is 120.

So, each of these lights will blink together after 2 minutes.
So, in an hour, they will blink 60/2 or 30 times.

Take the LCM of 100 cm and 85 cm = 1700 cm = 17 m

LCM of 18, 24 and 32 is 25 × 3² = 32 × 9 = 288
Hence, bells will chime together at an interval of 288 minutes or 4 hours and 48 minutes.