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EXERCISE 1 – NUMBER SYSTEMS
These questions will prove to be good exercise for some good practice in number
systems.
Try to do the questions without looking at the solutions. The time allotted is 30 mins.
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Question 1
The sum of
,
,
,…
is :
(1) less than 1
(2) greater than 1
(3) equal to 1 or less than 1
(4) equal to 1/n2 (n + 2)
The right answer is Option (1).
¼ + 1/9 + 1/16 +…+ 1/n2
Consider ¼:
¼ = 1/(2 x 2) < 1/(1 x 2) because ½ < 1
Similarly, 1/9 = 1/(3 x 3) < 1/(2 x 3)
Since 1/3 < ½
1/16 = 1/(4 x 4) < 1/(3 x 4)
Because ¼ < 1/3
……………………………………………………..
……………………………………………………..
1/n2 = 1/(nxn) < 1/(n – 1)n
Since 1/n < 1/(n – 1)
¼ + 1/9 + …+ 1/n2 < 1/(1 x 2) + 1/(2 x 3) + … + 1/[(n – 1)n]
i.e. ¼ + 1/9 + 1/16 + … + 1/n2 < (1/1 – ½) + (1/2 – 1/3) +…+ [1/(n – 1) – 1/n]
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i.e. ¼ + 1/9 + 1/16 +… + 1/n2 < 1/1 – 1/n which is less than 1.
Hence, the given sum is less than 1 only.
Answer: (1)
Question 2
What is the digit in the unit place of 3946252 + 734733 ?
(1) 8
(2) 7
(3) 6
(4) 4
The right answer is Option (1).
3946252 + 734733
First we will find the unit digit of 3946252.
Divide 46, by 4, we get 2 as remainder.
Actual remainder (2)252
Again dividing (2)252 by 4, we get 0 as remainder. And ‘9’ is an odd digit number. So, the last digit
of 3946252 will be 1.
Last digit of 734733  By dividing 47 by 4, we get 3 or – 1 as a remainder.
Now, (– 1) odd, when divided by 4 will give – 1 or 3 as remainder.
So, the last digit of 734733 will be.
Last digit of 3  3  3 = i.e. 7.
So, the last digit is 1 + 7 = 8. Answer: (1)
Question 3
How many terms, at the minimum of sequence 1,
for their sum to be not less than 3?
(1) 16
(2) 17
(3) 18
(4) 20
,
,
, …., must be added together
The right answer is Option (1).
1 + ½ + 1/3 + ¼ + 1/5 + 1/6 + 1/7 + …
Grouping the terms as follows
(1) + (1/2) + (1/3 + ¼) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + 1/10 + … 1/16) + …
(1/3 + ¼) > 2 x ¼, i.e. (1/3 + ¼) > ½
Similarly (1/5 + 1/6 + 1/7 + 1/8) > 4 x 1/8, i.e. > ½
Similarly (1/9 + 1/10 + … + 1/16) > 8 x 1/16,
i.e. > ½
If you take 16 terms their sum will be greater than 3. Answer: (1)
Question 4
When we divide 333435 by 34 the remainder is a and when we divide 333534 by 34 the
remainder is b. What is the value of b – a?
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(1) 0
(2) 32
(3) -32
(4) None of these
The right answer is Option (2).
333435 = (34 – 1)some even power =
333534 = (34 – 1)some odd power =
Now in first case, remainder a = 1
In second case, remainder b = 33
b – a = 33 – 1 = 32.
Answer: (2)
Question 5
ab and cd are two two-digit natural numbers. 4b + a = 13k1 and 5d – c = 17k2, where k1 and
k2 are natural numbers. The largest number that will always divide the product of ab and
cd is:
(1) 13
(2) 17
(3) 221
(4) None of these
The right answer is Option (3).
ab is always a multiple of 13 and cd is always a multiple of 17 so their product has to be divisible
by
17  13 = 221. Answer: (3)
Question 6
What is the number, the sum of whose factors is equal to
?
(1) 1124695
(2) 1123685
(3) 1012675
(4) None of these
The right answer is Option (1).
The sum of all the factor is given by
=
=
it means
So, the number is 132  113  51
 169  1331  5 = 1124695. Answer: (1)
Question 7
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A is the smallest integer which when multiplied with 3 gives a number made of 5’s only.
Sum of the digits of A is B. Sum of the digits of B is C. What is the value of C3?
(1) 125
(2) 64
(3) 216
(4) None of these
The right answer is Option (1).
Smallest number made of 5’s and divisible by 3 is 555. Thus A =
= 185.
Answer: (1)
Question 8
3762 – 2362 is divisible by:
(1) 560
(2) 840
(3) 775
(4) 160
The right answer is Option (2).
3762 – 2362
xn – yn is always divisible by (x – y)
xn – yn is divisible by (x + y) when n is even.
So, 3762 – 2362 will be divisible by both (37 – 23) and (37 + 23)
So, 3762 – 2362 will be divisible by (37 – 23) x
(37 + 23) = 14  60 = 840
Answer: (2)
Question 9
Find the smallest value of x such that x! ends with exactly 23 zeroes
(1) 100
(2) 98
(3) 99
(4) No such no. exist
The right answer is Option (4).
If we check that, 99! ends with how many zeroes, then we find that it ends with 22 zeroes
19 + 3 = 22
So, Now if we find that 100! ends with how many zeroes then it comes out be 24.
20 + 4 = 24
So, there are no such no. exists with ends with 23 zeroes. Answer: (4)
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Question 10
abc is a three-digit whole number so that abc = a3 + b3 + c3. [300 < abc < 400]. Find the sum
of a, b & c.
(1) 10
(2) 11
(3) 12
(4) Data insufficient
The right answer is Option (4).
Since 300 < abc < 400
 100a + 10b + c = a3 + b3 + c3
put a = 3 and rearrange, we get
300 + 10b + c = 27 + b3 + c3
 (b3 – 10b) + (c3 – c) = 273
Now put b = 6 and 7 (as b and c are both less than equal to 9 and 63 and 73 are the nearest
cubes to 273).
For b = 6. We get c3 – c = 117 and no value of c satisfies this.
For b = 7 we get c3 – c = 0 and two values of c i.e. 0 and 1.
Thus abc = 370 or 371
Hence, abc cannot be uniquely determined.
Answer: (4)
Question 11
Two different numbers when divided by the same divisor, left remainder 11 and 21
respectively, and when their sum was divided by the same divisor, then the remainder was
4. Find the divisor.
(1) 36
(2) 28
(3) 14
(4) 9
The right answer is Option (2).
Let the numbers are x1, x2 when divided by y (divisor), quotients are n & m and remainders are 11
& 21
respectively. Therefore
x1 = ny + 11
…..(1)
x2 = my + 21
…..(2)
Adding (1) and (2), x1 + x2 = (n + m)y + 32 = (n + m)y + 28 + 4
As the remainder given is 4, this is possible if y divides 28.
So y can be 14 or 28 but 14 is not possible because in (2) the remainder is 21. Answer : (2)
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Question 12
Which of the following numbers is divisible by 23?
(1) 22! + 1
(2) 22! – 1
(3) 21! + 2
(4) None of these
The right answer is Option (1).
According to the Wilson’s theorem
(P – 1)! + 1 is always divisible by P
Where P is a prime number.
So, we can say that
(23 – 1)! + 1 is divisible by 23
 22! + 1 is divisible by 23. Answer: (1)
Question 13
Let N = 103 + 104 + 105 + 106 + 107 + 108 + 109. The sum of the digits of N is:
(1) 12
(2) 1
(3) 6
(4) 7
The right answer is Option (4).
Since N is written out as a sum of powers of 10, then N can be written as 1111 111 000, and so
the sum of the digits is 7.
Answer: (4)
Question 14
If 27 = 123 and 31 = 133, then 15 = ?
(1) 13
(2) 31
(3) 11
(4) 33
The right answer is Option (4).
Let the base be N.
then 27 = (123)N
 27 = N2 + 2  N1 + 3  No = N2 + 2N + 3
 N2 + 2N – 24 = (N + 6) (N – 4) = 0  N = – 6, 4
But the base of a number system can’t be negative. Thus N = 4.
Now, 15 in base 4 can be written as (33)4.
Answer : (4)
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Question 15
The expression 333555 + 555333 is divisible by:
(1) 2
(2) 3
(3) 37
(4) All of these
The right answer is Option (4).
333555 + 555333
Check it by option.
The last digit of 333555 will be odd, and the last digit of 555333 will be odd.
odd + odd  even
So, the last digit is even, which is divisible by 2.
Both the numbers are individually divisible by 3. So 333555 + 555333 is divisible by 3.
And similarly, both the numbers are divisible by 37, so, 333555 + 555333 is divisible by 37.
Hence this expression is divisible by all three numbers. Answer : (4)
Note : If the number consists of same in the multiples of 3, then it is always a multiple of
37
Question 16
The value of (1.00008)12500 is
(1) < 1
(2) > 1
(3) > 2
(4) > 3
The right answer is Option (3).
(1.00008)12500 = (1 + .00008)12500 =
Applying binomial theorem: (a + b)n = nC0 an + nC1 an – 1 . b1 +……… + nCn – 1 . a1 . bn – 1 + nCn .
bn
= 1 + 12500 
= 2 + …………….+
+…………….+
> 2. Answer : (3)
Question 17
The last two digits of 2563  6325 are :
(1) 35
(2) 75
(3) 55
(4) 45
The right answer is Option (2).
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Last two digits of 2563 are always 25.
Last two digits of 6325 = an even no 3
(
Ten’s digit of any power of a number, whose unit digit is odd and Ten’s digit is even, is always
even).
Thus, last two digits of 2563  6325 = _ _ _ 2 5  an even no 3 = 75. Answer : (2)
Question 18
The remainder when 1010 + 10100 + 101000 + … + 1010000000000 is divided by 7 is:
(1) 0
(2) 1
(3) 2
(4) 5
The right answer is Option (4).
There are total of 10 terms.
Now apply congruence modulo theorem.
10  3 (mod 7)
10  – 4 (mod 7)
1010 = (–4)10 (mod 7)
= 410 (mod 7)
Now series can be re-written as.
410 + 4100 + 41000 + …………. + 410000000000
(43)3  4 + (43)33  4 + ……. + (43)333333333  4
Now 43  1 (mod 7)
Thus, remainder when above series is divided by 7 is
[(1)  4 + (1)  4 + ……. 10 times] = 40
40  5 (mod 7). Answer : (4)
Question 19
The number 6n2 + 6n for natural n is always divisible by
(1) 6 only
(2) 18
(3) 12 only
(4) 6 and 12
The right answer is Option (4).
Since the above expression contains a 6, it is always divisible by 6. Also, n (n + 1) is divisible by
2.
Hence, 6n (n + 1) is divisible by 6  2 = 12. The correct answer is 12 and 6.
Answer: (4)
Question 20
The number
is
(1) a finite number
(2) an infinite non–repeating decimal
(3) equal to 1.7320508
(4) an infinite repeating decimal
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The right answer is Option (2).
An infinite non–repeating decimal. Direct from definition. Answer: (2)
Question 21
Let D be a decimal of the form D = 0. abcd abcd abcd…………, where digits a, b, c and d
are integers lying between 0 and 9. At most three of these digits are zero. By what number
D be multiplied so that the result is a natural number?
(1) 999
(2) 9990
(3) 49995
(4) None of these
The right answer is Option (3).
D=
,
Also 49995 = 9999  5
Hence, once we multiply D with 49995 we will get a natural number.
Answer: (3)
Question 22
N = (11111111)2. What is the sum of the digits of N?
(1) 72
(2) 62
(3) 64
(4) None of these
The right answer is Option (3).
12 = 1, 112 = 121, 1112 = 12321
 111111112 = 123456787654321. The sum of these numbers is 64.
Answer: (3)
Question 23
What is the remainder when 192319241925 is divided by 1924?
(1) 1922
(2) 1923
(3) 1
(4) 1922
The right answer is Option (3).
192319241925 = (1924-1)some evenpower = x
 Remainder = (–1)x = 1.
Answer: (3)
Question 24
N = 2  4  6  8  10 ……. 100. How many zeroes are there at the end of N?
(1) 24
(2) 13
(3) 12
(4) None of these
The right answer is Option (3).
N = 250  50!. The highest power of 5 in 50! is 12, therefore there will be 12 zeroes at the end of
N. Answer: (3)
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Question 25
If in a number system, 40132 corresponds to 2542 in decimal, find the base of the number
system.
(1) 5
(2) 6
(3) 7
(4) 8
The right answer is Option (1).
Since, four is one of the characters in the given number, the base for that number system must
be > 5.
Assume that the base = 5. Then, (4  54) + (0  53) + (1  52) + (3  51) + (2  50) = 2542.
So, by estimation, we can say that the base of the number system is 5. Answer: (1)
Disclaimer: These are questions collected by me from various sources and I do not claim
any ownership to these. If you have any problems with them do contact me and I will
take steps to rectify them.
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