Test Level 2: Number System - 1 - CAT MCQ

Total number of pages = 200
Number of digits required to number the book from page 1 to 9 = 9
Number of digits required to number the book from page 10 to 99 = 2 × 90 = 180
Number of digits required to number the book from page 100 to 200 = 3 × 101 = 303
Therefore, total number of digits = 303 + 180 + 9 = 492

(i) X381 is the number given to us. If this number is divisible by 11, then the divisibility rule for 11 must be satisfied.
i.e. (X + 8) - (3 + 1) = (X + 4) is multiple of 11 and the smallest value of X for which X + 4 is divisible by 11 is 7. Hence, X = 7
(ii) 381Y is the number given to us. If this number is divisible by 9, then the divisibility rule for 9 must be satisfied.
i.e. (3 + 8 + 1 + Y) = Y + 12 must be divisible by 9. And smallest value of Y for which Y + 12 is divisible by 9 is 6. Hence, Y = 6

The expression is of the form p²X, where X = {q³(r - s) + t}.
So, whenever p is even, the product will be even, irrespective of whether the second number X is odd or even. So, p must be even.

Let the first number be x and the second number be y.
Then, x + 2y = 8 and x - y = 2
x + 2(x - 2) = 8
x + 2x - 4 = 8
3x = 12
x = 4
Since x - y = 2, this implies that y = 2.

If only weekend days are counted, only 2/7 portion of the whole week is calculated to arrive at her age.
So, 18 years of age = 2/7 (Actual age) ⇒ Actual age = 63 years

3¹ = 3, 3² = 9, 3³ = 27, 3⁴ = 81 and 3⁵ = 81 × 3 = 243.
Use the concept of power cycle (units digit repeats after the power of 4 and then after all powers that are multiples of 4).
Now, 99 = 24 × 4 + 3
Thus, 99 has 24 complete cycles of 4 and 3 is left as the remainder.
So, units digit of (173)⁹⁹ will be units digit of 3³, which is 7.

The required answer is nothing but the last digit of 8^13780000.
We know that, the cyclicity of 8 is 4. 
As 13780000 is divisible by 4 (cyclicity of 8).
That is, the last digit of 8⁴ = 6

1573 = 11 × 143
= 11 × 11 × 13
So, highest factor = 11 × 13 = 143

(ab)² = ccb
ccb > 300
The last digit of the number ab must be same as that of the square of ab.
So, b can be 0, 1, 5 or 6.
20² =400 and 30² =900 are three digit numbers and greater than 300. But the first 2 digits are not same. Hence, b is not 0.

If b is 5, then the ten's digit of ab's square will be 2 => c = 2. But if c is 2, then ccb is not greater than 300. Hence, b is not 5.

If b is 6, then 26² = 576 is the only three digit number that is greater than 300. But, it is not in the form of ccb => b is not 6.

If b is 1, then 21² =441 satisfies all the given conditions => b is 1
 

It is clear that for n positive integers function h (a,b) has to be used one time less than the number of integers, i.e., (n-1) times.