Test Level 3: Number System - 1 - CAT MCQ

A * B = (A + B)(A × B - A² + B²)
7 * 5 = (7 + 5)(35 - 7² + 5²)
= (12)(35 - 49 + 25)
= 132

34300 = 7 × 7 × 7 × 5 × 5 × 2 × 2 = 7³ × 5² × 2²
To make 34300 a perfect square, multiply it by 7.

There will be a cycle of last digits, i.e. 8, 4, 2, 6,... and this cycle consists of 4 numbers.
To find the last digit, divide the required power by 4 and find the remainder.
So, when 12,500 is divided by 4, it gives 0 as the remainder.
Therefore, the last digit required would be 6.

33 = 3 × 11
44 = 2² × 11
175 = 5² × 7
80 = 2⁴ × 5
180 = 5 × 2² × 3²
66 = 2 × 3 × 11
12 = 2² × 3
Multiplying all, we get
2¹¹ × 5⁴ × 3⁵ × 7 × 11³ = 10⁴ × 2⁷ × 3⁵ × 7 × 11³
Look out for the number of tens, making total number of zeros.
Number of zeros = 4

1260 = 4 × 315
= 4 × 5 × 63
= 2² × 3² × 5 × 7
The factor has to end with a zero means it should have 2 and 5 necessarily.
So, if we take out 2 and 5, then we will have factors that not ended with 0 - 2 × 3² × 7.
The required answer is the number of factors of 2 × 3² × 7.
⇒ (1 + 1)(2 + 1)(1 + 1)
= 12

Sum of the digits at even places = 11 + x
Sum of the digits at odd places = 11 + x
Difference is 0.
i.e. The given number is divisible by 11, irrespective of the value of x.
So, x can be anything from 0 to 9.
Thus, x can be substituted with 10 values.

Suppose that the last two digits of the four-digit number are c and d, where c is at the tens place and d is at the units place.
We can ignore the first two digits.
This gives us,
(10c + d) - (10d + c) = 54
9(c - d) = 54
c - d = 6

If n = 1, 2¹ is divisible by 1²
n = 2, 2² is divisible by 2²
n = 4, 2⁴ is divisible by 4²
n = 8, 2⁸ is divisible by 8²
n = 16, 2¹⁶ is divisible by 16²
And so on...
So, 'n' can take infinite values, such that 2ⁿ is exactly divisible by n².

Let the hundreds digit be a.
Then, the tens digit = a + 1 and the units digit = a + 2
⇒ The number = 100a + 10(a + 1) + a + 2 = 111a + 12
The number formed by reversing the digits = 100(a + 2) + 10(a + 1) + a
= 111a + 210 
Since the number formed by reversing the digits exceeds the original number by 22 times, the sum of the digits
⇒ 111a + 210 - 111a - 12 = 22(a + 2 + a + 1 + a)
⇒ 198 = 66a + 66 
⇒ a = 2
Hence, the required number = 111a + 12 = 111 × 2 + 12 = 234

Only 1,43,616 is a multiple of both 11 and 8 and so of 88.
For a number to be divisible by '8' the last three digits should be multiple of '8'.
Option (1) 13,92,578 the last three digits i.e. 578 is not divisible by 8.
Option (2) 1,38,204 the last three digits i.e. 204 is not divisible by 8.
Option (3) 14,36,240 the last three digits i.e. 240 is divisible by 8.
Option (4) 1,43,616 the last three digits i.e. 616 is divisible by 8.
Option (5) 1,43,661 the last three digits i.e. 661 is not divisible by 8.
Now check option (3) and option (4) for multiple of '11'.
For 14,36,280 the sum of digits at odd places and sum of digits at even places is 0 + 2 + 3 + 1 = 6 and 8 + 6 + 4 = 18.
Now 18 - 6 = 12 not multiple of 11. Hence, number 14,36,280 is not multiple of 88.
For 1,43,616 the sum of digits at odd places and sum of digits at even places is 6 + 6 + 4 = 16 and 1 +3 + 1 = 5.
Now 16 -5 = 11.
Hence, 1,43,616 is multiple of 88.