Previous Year Topic Wise Questions With Solutions: Number System | CSAT Preparation - UPSC PDF Download

Ans: (b)
Let's look at the pattern of squares of numbers consisting only of ones: 
1² = 1 (1 digit in N, 1 digit in N², highest digit in N² is 1)
11² = 121 (2 digits in N, 3 digits in N², highest digit in N² is 2) 
111² = 12321 (3 digits in N, 5 digits in N², highest digit in N² is 3) 
1111² = 1234321 (4 digits in N, 7 digits in N², highest digit in N² is 4) 
11111² = 123454321 (5 digits in N, 9 digits in N², highest digit in N² is 5) …….and so on. 
In our case, N²=12345678987654321. The highest digit in the sequence is 9, and the number of digits in it is 17. This means that N must be a number consisting of nine '1's. So, N = 111,111,111.
Thus, the number N has 9 digits.

Ans: (c)
Method I: The given expression: 1 × 3 × 5 × 7 × 9 × … × 999 1 × 3 × 5 × 7 × 9 = 945
So, the unit digit is 5. Similarly, the unit digit of the product 11 × 13 × 15 × 17 × 19 would also be 5. 
Same would be the case for each such set, i.e.(21 × 23 × 25 × 27 × 29),  ………. 991 × 993 × 995 × 997 × 999. 
So, basically, we will get 5 × 5 × 5 × ….. × 5 Unit digit of the product 5 × 5 × 5 × ….. × 5 would also be 5. 
Method II: The given expression: 1 × 3 × 5 × 7 × 9 × … × 999 We can see that there is 5 in the multiplication. 
Multiplication rule of 5: (i) If you multiply an odd number by 5, the product will always end in 5.
Example: (3 × 5 = 15), (7 × 5 = 35), ….. etc. 
(ii) If you multiply an even number by 5, the product will always end in 0.
Example: (4 × 5 = 20), (6 × 5 = 30), ….. etc. 
In the given expression, only odd numbers are being multiplied. 
So, the final product's unit digit must be 5.

Ans: (c)
If the numbers are 1 and 11, there are two numbers between them that are divisible by 5 (i.e. 5 and 10). 
However, if we consider the numbers 5 and 15, there is only one number between them that is divisible by 5 (i.e. 10). 
Hence, there can be one or two such numbers in the given range, depending on the range we choose.

Ans: (c)
x, y and z are distinct prime numbers less than 10. So, these numbers can be 2, 3, 5, or 7.
S = x + y + z
S = 2 + 3 + 5 = 10..................(1)
S = 3 + 5 + 7 = 15...................(3)
So, (2,3,5) and (3,5,7) satisfy conditions 1 and 3 respectively. 
However, statement 2 is not true for these given set of numbers.

Ans: (c)
32⁵ + 2²⁷ = (2⁵)⁵ + 2²⁷ (∵ (a^m)ⁿ = a^mn) 
= 2²⁵ + 2²⁷
= 2²⁵(1 + 2²) 
= 2²⁵ × 5 
= 2^{24 + 1} × 5 
= 2²⁴ × 2¹ × 5 (∵ a^m+n = a^m × aⁿ)
= 2²⁴ × 10
Thus, 32⁵ + 2²⁷ is completely divisible by 10.

Ans: (a)
​Here, we are just adding up the numbers and then summing up the digits of the resultant number. 
7 ⨁ 9 ⨁ 10 = 7 + 9 + 10 = 26 = 2 + 6 = 8
9 ⨁ 11 ⨁ 30 = 9 + 11 + 30 = 50 = 5 + 0 = 5
11 ⨁ 17 ⨁ 21 = 11 + 17 + 21 = 49 = 4 + 9 = 13
So, 23 ⨁ 4 ⨁ 15 = 23 + 4 + 15 = 42 = 4 + 2 = 6
Hence, option (a) is correct.

Ans: (a)
N = 9999 …… 99 times 
Any digit repeated (P - 1) times is divisible by P, where P is a prime number > 5. 
So, 9999 …… repeated 13 - 1 = 12 times will be divisible by 13. 
So, 9999 …… repeated 12 × 8 = 96 times will be divisible by 13. 
That is, Remainder [9999...... 96 times / 13] = 0 
Or Remainder [9999...... (96 times) 000 / 13] = 0
So, we just need to find out the remainder when we divide the remaining three digits by 13.
Remainder [999 / 13] = 11 
Hence, option (a) is correct. 
Method II: We can analyze the pattern of remainders. 
Remainder [9/13] = 9 
Remainder [99/13] = 8 
Remainder [999/13] = 11 
Remainder [9999/13] = 2 
Remainder [99999/13] = 3 
Remainder [999999/13] = 0 
This pattern can be seen getting repeated thereafter too. 
Remainder [9999999/13] = 9 
Remainder [99999999/13] = 8 …and so forth.
So, if total number of 9s is six, twelve, eighteen, ……., ninety, ninety six, etc., the remainder is 0.
So, if the number has ninety seven 9s, the remainder is 9. [Following the pattern] 
So, if the number has ninety eight 9s, the remainder is 8. 
So, if the number has ninety nine 9s, the remainder is 11. 
Therefore, the answer is 11.
Hence, option (a) is correct.

Ans: (c)
Method I: 11 × 11 = 121
111 × 111 = 12321
1111 × 1111 = 1234321
Following the same pattern, we get: 
(1111 …. 9 times) × (1111 …. 9 times) = 12345678987654321 
Sum of the digits of the resulting number = 2 × (1 + 2 + 3 + …. + 8) + 9 = 2 × [8 × 9/2] + 9 = 72 + 9 = 81
[Sum of first n natural numbers = n (n + 1) / 2] Hence, option (c) is correct. 
Method II: Since each digit of a 9-digit number is 1, so the sum of its digits = 9. 
So, this number is divisible by 9. Any multiple of such a number will also be divisible by 9. 
So, the sum of the digits of the resulting number of the multiplication (111111111) × (111111111) must also be divisible by 9.
Therefore, the correct answer will be the option which is a multiple of 9, i.e. 81.
Hence, option (c) is correct.

Ans: (a)
It’s given that, ABC × D = 37DD Wherein A, B, C and D are different non-zero digits. 
So, ABC = 37DD / D = (3700 + 10D + D) / D = (3700/D) + 11
The possible values of D, such that ABC is an integer, are 1, 2, 4, and 5. 
If D = 1, ABC = 3700 + 11 = 3711. It can be rejected as ABC is a three-digit number.
If D = 2, ABC = 1850 + 11 = 1861. It can be rejected as ABC is a three-digit number.
If D = 5, ABC = 740 + 11 = 751. It can be rejected as here B = D = 5. 
If D = 4, ABC = 925 + 11 = 936. 
So, A + B + C = 9 + 3 + 6 = 18 
Hence, option (a) is correct.

Ans: (d)
Divisor = Dividend × Quotient + Remainder 
So, 1186 = Dividend × Natural number + 31 
Or Dividend × Natural number = 1186 – 31 = 1155
Hence, the required natural number must be a factor of 1155.
Now, 1155 = 1 × 3 × 5 × 7 × 11
When 1186 is divided by the required natural number, we get 31 as remainder. 
Thus, the required natural number must be greater than 31. 
Factors of 1155 which are greater than 31 are: 33, 35, 55, 77, 105, 165, 231, 385, and 1155. 
Thus, there are 9 natural numbers which give a remainder of 31 when 1186 is divided by them.
Hence, option (d) is correct.

Ans: (a)
The number has 7 digits, and has been denoted by: ABCDEFG 
These letters can be replaced by 1, 2, 4, 5, 7, 8, 9, not necessarily in the same order. 
We have to find the possible value of C + D + E The original number (ABCDEFG) is divisible by 9. 
It has to be as 1 + 2 + 4 + 5 + 7 + 8 + 9 = 36, which is divisible by 9. This information is utterly useless.
After deleting 1 digit from the right, the resulting number (ABCDEF) is divisible by 6. 
It means that, F = 2, 4 or 8 (i.e. an even number). 
Also, if even after removing G, the remaining number is divisible by 3, then it means G = 9.
After deleting 3 digits from the right, the resulting number (ABCD) is divisible by 4. 
It means that, D = 2, 4 or 8 (i.e. an even number).
After deleting 5 digits from the right, the resulting number (AB) is divisible by 2. 
It means that, B = 2, 4 or 8 (i.e. an even number).
So, F, D and B are even numbers (2, 4 or 8). And, A, C, E, and G are odd numbers (1, 5, 7 or 9).
After deleting 2 digits from the right, the resulting number (ABCDE) is divisible by 5. It means that, E = 5.
So, we just have to find C + D + E = C + D + 5, which must be an even number as C is odd (1, or 7), and D is even (2, 4, or 8).
On observing the options, we can see that C must be 1 and D must be 2. So, C + D + E = 1 + 2 + 5 = 8.

Ans: (b)
We need to find three-digit numbers in which: 
All digits are different, and all digits are odd. So, the three digits must be from amongst 1, 3, 5, 7, and 9. 
The number is divisible by 5, i.e. the units digit is 5. 
The number of ways we can fill the first two digits from amongst 4 distinct digits = 4 × 3 = 12.

Ans: (c)
LCM of 6, 9, 12, 15 and 18 = 180
Smallest number greater than 1000 which is a multiple of 180 is 1080.
So, Required number = 1080 + 3 = 1083

Ans: (d)
p is a two-digit number, and q is the number consisting of the same digits in the reverse order.
It’s given that, p × q = 2430 
The last digit of the product is 0, which indicates that one two of the digits must be 5.
Let the remaining digit be x.
So, p = x5 = x × 10 + 5 = 10x + 5 
Reverse number, q = 5x = 5 × 10 + x = 50 + x 
According to the question, (10x + 5) × (50 + x) = 2430 
Or 500x + 10x² + 250 + 5x = 2430 
Or 10x² + 505x – 2180 = 0 
Or 2x² + 101x – 436 = 0 
Or x(2x + 109) – 4(2x + 109) = 0 
Or (x – 4)(2x + 109) = 0 
Or x = 4, or –109/2 (can be neglected) 
So, x = 4 
Therefore, p = x5 = 45 
And q = 54 
Required difference = 54 – 45 = 9

Ans: (a)
Since, X and Y are distinct non-zero digits 
∴ The required number of numbers of the form 0.XY = 9 × 8 = 72 
Hence, option (a) is the correct answer.

Ans: (d)
Here we have to find out all the integers between 700 to 1000, in which sum of the digits is 10, e.g. 703 ⟶ 7 + 0 + 3 = 10
These numbers have been listed below: 703, 712, 721, 730, 802, 811, 820, 901 and 910 
Hence, there are 9 such integers in which the sum of the digits is 10.
Hence, option (d) is the correct answer.

Ans: (c)
It’s given that: 3²⁰¹⁹ is divided by 10. Now, 31 = 3 32 = 9
33 = 27
34 = 81
35 = 243
36 = 729
Since, unit place of the power of 3 repeats after every 4 steps (i.e. it has a cyclicity of 4). 
Now, on dividing 2019 by 4 we get a remainder of 3. 
Hence, 3²⁰¹⁹ will have the same last digit as that of 33, i.e. 7. 
(33)/10 = 27/10 
Hence, the remainder will be 7.

Ans: (b)
Given:  3798125P369 is divisible by 7 
Let’s express the given number in terms of triplets of digits, starting from the right, as follows. (37) (981) (25P) (369)  
Now, 369 – 25P + 981 – 37 = 1350 – 25P – 37 = 1313 – 25P 
By dividing 1313 by 7 we get 4 as remainder. 
25P in the above equation should be in between 250 to 259. 
Now, 252 and 259 are divisible by 7. 
So, the number must be 252 + 4 = 256. 
Hence, value of P = 6 
Let’s double check it: 1313 – 256 = 1057
It is divisible by 7.
Hence, option (b) is the correct answer.

Ans: (d)
Method I: As the number comprising of all 1’s is obtained on multiplication by 7, so it means that 7 is the factor of that number. 
Our answer will the smallest number comprising of all 1’s that will be divisible by 7. So, let’s check.
Is 1 divisible by 7? – No 
Is 11 divisible by 7? – No 
Is 111 divisible by 7? – No 
Is 1111 divisible by 7? – No 
Is 11111 divisible by 7? – No 
Is 111111 divisible by 7? – Yes 
So, 111111/7 = 15,873 
Method II: Multiply the numbers given in the options by 7, and see which multiplication gives you a resultant comprising of all 1’s. 
Checking option (d): 15873 × 7 = 111111

Ans: (b)
For the number to be greater than 30000, it must start with the digit 3. Also, as only 5 digits are given to us, all must be used. 
3 _ _ _ _

The 4 blanks have to be filled by two 2’s and two 3’s. 
Number of ways to do so = 4!/(2! 2!) = 6 
These numbers are: 33322, 33232, 33223, 32332, 32323, and 32233.

Ans: (c)
3P + 4P + PP + PP = RQ2
Or 30 + P + 40 + P + 10P + P + 10P + P = 100R + 10 Q + 2 
Or 24P + 70 = 100R + 10 Q + 2 
Or 20P + 70 + 4P = 100R + 10 Q + 2 
The unit digit of the resultant is 2. It will be obtained when 4 is multiplied by P. 
So, P must be 3, or 8. If P = 3, then: 24P + 70 = 24 × 3 + 70 = 72 + 70 = 142
If P = 8, then: 24P + 70 = 24 × 8 + 70 = 192 + 70 = 262
Arithmetic sum of 142 and 262 = (142 + 262)/2 = 202