## Elitmus Previous Years Questions On Number System

### SET 1

# Number System 02

01. | If n is any odd number greater than 1, then n(n^{2 }- 1) is always divisible by ? | |||

(a) 96 | (b) 48 | (c) 24 | (d) None of these |

02. | If a number 774958A96B is to be divisible by 8 and 9, the respective values of A and B will be | |||

(a) 7 and 8 | (b) 8 and 0 | (c) 5 and 8 | (d) None of these |

03. | Three consecutive positive even numbers are such that thrice the first number exceeds double the third by 2, the third number is ? | |||

(a) 10 | (b) 14 | (c) 16 | (d) 12 |

04. | Three bells chime at intervals of 18 min, 24 min and 32 min respectively. At a certain time, they begin to together, What length of time will elapse before they chime together again ? | |||

(a) 2 h and 24 min | (b) 4h and 48 min | (c) 1 h and 36 min | (d) 5h |

05. | Two positive integers differ by 4 and sum of their reciprocals is 10/21. Then, one of the numbers is ? | |||

(a) 3 | (b) 1 | (c) 5 | (d) 21 |

06. | (5^{6}-1) is divisible by ? | |||

(a) 13 | (b) 31 | (c) 5 | (d) None of these |

07. | The remainder obtained when a prime number greater that 6 is divided by 6 is | |||

(a) 1 or 3 | (b) 1 or 5 | (c) 3 or 5 | (d) 4 or 5 |

08. | For the product n(n+1)(2n+1) , which one of the following is not necessarily true ? | |||

(a) It is even | (b) Divisible by 3 | (c) divisible by (n(n+1)(2n+1))/2 | (d) Never divisible by 237 |

09. | 72 Hens Rs …96.7.. Then what does each hen cost, where two digits in place of “…. , …..”are not visible written in illegible hand-writing ? | |||

(a) Rs 3.23 | (b) Rs 5.11 | (c) Rs 5.51 | (d) Rs 7.22 |

10. | Which is the least number that must be subtracted from 1856 so that the remainder when divided by 7, 12, 16, is 4? | |||

(a) 137 | (b) 1361 | (c) 140 | (d) 172 |

Answers : Find Detailed Solutions at the end of the the page.

1. | C | 2. | B |

3. | B | 4. | B |

5. | A | 6. | B |

7. | B | 8. | D |

9. | C | 10. | D |

### SET 2

# Number System 01

###
01. The number of common terms in the two sequences 17, 21, 25, ….. , 417 and 16, 21, 26, …,466 is ?
(a) 19 (b) 20 (c) 77 (d) 22

02. How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed ??
(a) 374 (b) 500 (c) 375 (d) 376

03. What is the number of distinct terms in the expansion of (a + b + c)^{20} ?
(a) 231 (b) 253 (c) 242 (d) 210

04. What are the last two digits of 7^{2008} ?
(a) 21 (b) 61 (c) 01 (d) 41

05. The integers 1, 2, 3, … 40 are written on blackboard. The following operation is then repeated 39 times. In each repetition, any two numbers, say a and b, currently on the blackboard are erased and a new number a + b – 1 is written. What will be the number left on the board at the end ?
(a) 820 (b) 821 (c) 781 (d) 819

06. An intelligence agency decides on a code of 2 digits selected from 0, 1, 2, …, 9. But on the slip on which the code is hand written allows confusion between top and bottom, because there are indistinguishable. Thus, for example, the code 91 could be confused with 16. How many codes are there such that there is no possibility of any confusion?
(a) 25 (b) 75 (c) 80 (d) None of these

07. A young girl counted in the following way on the fingers of her left hand. She started calling the thumb 1, the index finger 2, middle finger 3, ring finger 4, little finger 5, then reversed direction, calling the ring finger 6, middle finger 7, index finger 8 and thumb 9 and then back to the index finger for 10, middle finder for 11 and so on. She counted up to 1994. She ended on her ?
(a) thumb (b) index finger (c) middle finger (d) ring finger

08. Let U(n+1)= 2Un + 1, ( n=0, 1, 2, …………) U0=0 then U(10) would be nearest to ?
(a) 1023 (b) 2047 (c) 4095 (d) 8195

09. The product of all integers from 1 to 100 will have the following numbers of zeros at the end ?
(a) 20 (b) 24 (c) 19 (d) 22

10. The number of positive integers not greater than 100, which are not divisible by 2, 3 or 5 is ?
(a) 26 (b) 18 (c) 31 (d) None of these

Answers : Find Detailed Solutions at the end of the the page.
1. B 2. D
3. A 4. C
5. C 6. C
7. B 8. A
9. B 10. A

01. | The number of common terms in the two sequences 17, 21, 25, ….. , 417 and 16, 21, 26, …,466 is ? | |||

(a) 19 | (b) 20 | (c) 77 | (d) 22 |

02. | How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed ?? | |||

(a) 374 | (b) 500 | (c) 375 | (d) 376 |

03. | What is the number of distinct terms in the expansion of (a + b + c)^{20} ? | |||

(a) 231 | (b) 253 | (c) 242 | (d) 210 |

04. | What are the last two digits of 7^{2008} ? | |||

(a) 21 | (b) 61 | (c) 01 | (d) 41 |

05. | The integers 1, 2, 3, … 40 are written on blackboard. The following operation is then repeated 39 times. In each repetition, any two numbers, say a and b, currently on the blackboard are erased and a new number a + b – 1 is written. What will be the number left on the board at the end ? | |||

(a) 820 | (b) 821 | (c) 781 | (d) 819 |

06. | An intelligence agency decides on a code of 2 digits selected from 0, 1, 2, …, 9. But on the slip on which the code is hand written allows confusion between top and bottom, because there are indistinguishable. Thus, for example, the code 91 could be confused with 16. How many codes are there such that there is no possibility of any confusion? | |||

(a) 25 | (b) 75 | (c) 80 | (d) None of these |

07. | A young girl counted in the following way on the fingers of her left hand. She started calling the thumb 1, the index finger 2, middle finger 3, ring finger 4, little finger 5, then reversed direction, calling the ring finger 6, middle finger 7, index finger 8 and thumb 9 and then back to the index finger for 10, middle finder for 11 and so on. She counted up to 1994. She ended on her ? | |||

(a) thumb | (b) index finger | (c) middle finger | (d) ring finger |

08. | Let U(n+1)= 2Un + 1, ( n=0, 1, 2, …………) U0=0 then U(10) would be nearest to ? | |||

(a) 1023 | (b) 2047 | (c) 4095 | (d) 8195 |

09. | The product of all integers from 1 to 100 will have the following numbers of zeros at the end ? | |||

(a) 20 | (b) 24 | (c) 19 | (d) 22 |

10. | The number of positive integers not greater than 100, which are not divisible by 2, 3 or 5 is ? | |||

(a) 26 | (b) 18 | (c) 31 | (d) None of these |

Answers : Find Detailed Solutions at the end of the the page.

1. | B | 2. | D |

3. | A | 4. | C |

5. | C | 6. | C |

7. | B | 8. | A |

9. | B | 10. | A |

# Number System Solution 01

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1.

Both the sequences ( 17, 21, 25 ……………… and (16, 21, 26….. are arithmetic progression with a common difference of 4 and 5 respectively.

In both the sequence first common term is 21.

Hence a new arithmetic sequence containing the common terms of both the series can be formed with a common difference of LCM of ( 4, 5) is 20

New sequence will be 21, 41, 61, ….401

n^{th }term = a + (n-1)d

401= 21 + (n-1)20

n-1=19

Hence, n=20

(b)

2.

The number required is greater than 999 and less than and equal to 4000.

Now out off our digits, 0, 1, 2, 3, 4.

To form a number greater than 999 and less than 4000.

The digit at thousands place can be selected in 3 ways( 0 and 4 cannot be taken )

The digit at hundreds place can be selected in 5 ways.

The digit at tens place can be selected in 5 ways.

Total required number of ways= 3 x 5 x 5 x 5 = 375 ways

Since, 4000 is also one of the required number.

Therefore, total number of ways=375 + 1

(d)

5.

According to question, if two numbers say a and b are erased and replaced by a new number a + b – 1, then in every repetition, the number of integers gets reduced by 1 and consequently at the last repetition there will be only one number left.

Whatever may be our selection of two numbers a and b. In any and every repetition, the final number so arrived will not changes.

Now, the sum of integers from 1 to 40= n(n + 1)/2 = 820

As, discussed above the sum of integers of the first, second, third ……… repetitions will be 819, 818, 817, ………….. so on respectively. Therefore, after 39 operations there will be only 1 number left and that will be 820 – 39=781

6.

(c)

7.

Thumb Finger :1, 9, 17, ….

Index Finger: 2, 8, 10, 16, 18, ….

Middle Finger: 3, 7, 11, 16, 19, ….

Ring Finger : 4, 6, 12, 14, 20, ……

Little Finger : 5, 13, 21, ….

Numbers on thumb forms a AP with common difference=8

Numbers on middle forms a AP with common difference=4

1993 will be on thumb.

Hence, 1994 will be on index finger.

8.

U(n+1)= 2Un + 1, ( n=0, 1, 2, …………) U0=0

Put

n=0, U1=1

n=1, U2=3

n=2, U3=7

n=4, U5=31

Seeing this pattern we can conclude i.e. U(n)= 2^{n} -1

Hence U(10)= (2)^{10 }- 1= 1023

9.

Every combination of 5 and 2 will give one zero, and number of zero in the product of any number is decided by the number of 2 and 5, whichever is less.

Hence, this problem can be solved by determining the number of 2 and 5 between 1 to 100.

Clearly there are 20 numbers which are divisible by 5. Besides, there are four numbers 25, 50, 75, and 100 which will have one addition 5. Hence, number of zeroes in the product of all the numbers from 1 to 100 is 21.

(b)

10.

There are 50 odd numbers less than 100 which are not divisible by 2. Out of these 50 there are 17 number which are divisible by 3.

Out of remaining there are 7 numbers which are divisible by 5.

Hence, numbers which are not divisible by 2, 3, 5 = 50 – 17 – 7 = 26

1. Both the sequences ( 17, 21, 25 ……………… and (16, 21, 26….. are arithmetic progression with a common difference of 4 and 5 respectively. In both the sequence first common term is 21. Hence a new arithmetic sequence containing the common terms of both the series can be formed with a common difference of LCM of ( 4, 5) is 20 New sequence will be 21, 41, 61, ….401 n ^{th }term = a + (n-1)d401= 21 + (n-1)20 n-1=19 Hence, n=20 (b) |

2. The number required is greater than 999 and less than and equal to 4000. Now out off our digits, 0, 1, 2, 3, 4. To form a number greater than 999 and less than 4000. The digit at thousands place can be selected in 3 ways( 0 and 4 cannot be taken ) The digit at hundreds place can be selected in 5 ways. The digit at tens place can be selected in 5 ways. Total required number of ways= 3 x 5 x 5 x 5 = 375 ways Since, 4000 is also one of the required number. Therefore, total number of ways=375 + 1 (d) |

5. According to question, if two numbers say a and b are erased and replaced by a new number a + b – 1, then in every repetition, the number of integers gets reduced by 1 and consequently at the last repetition there will be only one number left. Whatever may be our selection of two numbers a and b. In any and every repetition, the final number so arrived will not changes. Now, the sum of integers from 1 to 40= n(n + 1)/2 = 820 As, discussed above the sum of integers of the first, second, third ……… repetitions will be 819, 818, 817, ………….. so on respectively. Therefore, after 39 operations there will be only 1 number left and that will be 820 – 39=781 |

6. (c) |

7. Thumb Finger :1, 9, 17, …. Index Finger: 2, 8, 10, 16, 18, …. Middle Finger: 3, 7, 11, 16, 19, …. Ring Finger : 4, 6, 12, 14, 20, …… Little Finger : 5, 13, 21, …. Numbers on thumb forms a AP with common difference=8 Numbers on middle forms a AP with common difference=4 1993 will be on thumb. Hence, 1994 will be on index finger. |

8. U(n+1)= 2Un + 1, ( n=0, 1, 2, …………) U0=0 Put n=0, U1=1 n=1, U2=3 n=2, U3=7 n=4, U5=31 Seeing this pattern we can conclude i.e. U(n)= 2 ^{n} -1Hence U(10)= (2) ^{10 }- 1= 1023 |

9. Every combination of 5 and 2 will give one zero, and number of zero in the product of any number is decided by the number of 2 and 5, whichever is less. Hence, this problem can be solved by determining the number of 2 and 5 between 1 to 100. Clearly there are 20 numbers which are divisible by 5. Besides, there are four numbers 25, 50, 75, and 100 which will have one addition 5. Hence, number of zeroes in the product of all the numbers from 1 to 100 is 21. (b) |

10. There are 50 odd numbers less than 100 which are not divisible by 2. Out of these 50 there are 17 number which are divisible by 3. Out of remaining there are 7 numbers which are divisible by 5. Hence, numbers which are not divisible by 2, 3, 5 = 50 – 17 – 7 = 26 |

# Number System Solution 02

###
1.

We have use put different values of n ( odd numbers ) greater than 1.

i.e. n=3, 5, 7, 9

When n=3 n(n^{2 }- 1)= 24

When n=5 n(n^{2 }- 1)= 120

When n=7 n(n^{2 }- 1)=336

using options we find that all the numbers are divisible by 24

(c)

2.

Numbers are **divisible by 8** if the number formed by the last three digits is evenly divisible by 8.

Numbers are **divisible by 9** if the sum of all the individual digit is divisible by 9.

The number 774958A96B is divisible by 8 if 96B is divisible by 8. And 96B is divisible by 8 if B is either 0 and 8. Now to make the same number divisible by 9 sum of all the digits should be divisible by 9. Hence, (55 + A + B) is divisible by 9 if (A + B) is either 0 or 8

i.e. either A = 0 or B = 8 or A = 8 or B = 0

Since, the number is divisible by both A and B. Hence, A and B may take either values i.e. 8 and 0

(b)

3.

Let the three even number are ( x – 2), x, (x + 2)

Then, 3(x – 2) – 2(x + 2) = 2

3x – 6 – 2x – 4 = 2 i.e. x=12

Hence, the third number is (12 + 2) = 14

(b)

4.

We have to take the L.C.M of 18, 24 and 32 i.e. 288 min.

Hence, bells will chime together again after 4 Hours and 48 Minutes.

(b)

6.

(5^{6 }- 1) = (5^{3})^{2 }- (1)^{2 }= (125)^{2 }- (1)^{2}

=(125 + 1)(125 – 1) = 126 x 124=31 x 2 x 2 x 126

Here we can easily conclude. ( 5^{6}-1) is divisible by 31

(b)

7.

We have take some prime number greater than 6 i.e. 7, 11, 13, 17, 19, 23, 29,31, 37, 41

Now we have divide the numbers by 6. The remainder is always either 1 or 5.

(b)

8.

We have to check for each option separately by taking the different values of n.

Option (a) : Check for n= 3, 4, 5, 6…

Option (b) : Check for n= 3, 4, 5, 7…

Option (c) : Divisible

Option (d) : For n= 237 : n(n + 1)(2n + 1) is divisible.

(d)

9.

We have to check for each option.

(a) 3.23 x 72 = 232.56

(b) 5.11 x 72 = 367.92

(c) 5.51 x 72 = 396.72

(d) 7.22 x 72 =519.84

Option (c) is nearest to Rs. …..96.7…

(c)

10.

Firstly, we have to take LCM of 7, 12, 16 = 336

If we divide 1856 by 336, then remainder is 176. Since it is given that remainder in this condition is 4. Hence, the lease number to be subtracted = (176 – 4)172.

(d)

1. We have use put different values of n ( odd numbers ) greater than 1. i.e. n=3, 5, 7, 9 When n=3 n(n ^{2 }- 1)= 24When n=5 n(n ^{2 }- 1)= 120When n=7 n(n ^{2 }- 1)=336using options we find that all the numbers are divisible by 24 (c) |

2. Numbers are divisible by 8 if the number formed by the last three digits is evenly divisible by 8.Numbers are divisible by 9 if the sum of all the individual digit is divisible by 9.The number 774958A96B is divisible by 8 if 96B is divisible by 8. And 96B is divisible by 8 if B is either 0 and 8. Now to make the same number divisible by 9 sum of all the digits should be divisible by 9. Hence, (55 + A + B) is divisible by 9 if (A + B) is either 0 or 8 i.e. either A = 0 or B = 8 or A = 8 or B = 0 Since, the number is divisible by both A and B. Hence, A and B may take either values i.e. 8 and 0 (b) |

3. Let the three even number are ( x – 2), x, (x + 2) Then, 3(x – 2) – 2(x + 2) = 2 3x – 6 – 2x – 4 = 2 i.e. x=12 Hence, the third number is (12 + 2) = 14 (b) |

4. We have to take the L.C.M of 18, 24 and 32 i.e. 288 min. Hence, bells will chime together again after 4 Hours and 48 Minutes. (b) |

6. (5 ^{6 }- 1) = (5^{3})^{2 }- (1)^{2 }= (125)^{2 }- (1)^{2}=(125 + 1)(125 – 1) = 126 x 124=31 x 2 x 2 x 126 Here we can easily conclude. ( 5 ^{6}-1) is divisible by 31(b) |

7. We have take some prime number greater than 6 i.e. 7, 11, 13, 17, 19, 23, 29,31, 37, 41 Now we have divide the numbers by 6. The remainder is always either 1 or 5. (b) |

8. We have to check for each option separately by taking the different values of n. Option (a) : Check for n= 3, 4, 5, 6… Option (b) : Check for n= 3, 4, 5, 7… Option (c) : Divisible Option (d) : For n= 237 : n(n + 1)(2n + 1) is divisible. (d) |

9. We have to check for each option. (a) 3.23 x 72 = 232.56 (b) 5.11 x 72 = 367.92 (c) 5.51 x 72 = 396.72 (d) 7.22 x 72 =519.84 Option (c) is nearest to Rs. …..96.7… (c) |

10. Firstly, we have to take LCM of 7, 12, 16 = 336 If we divide 1856 by 336, then remainder is 176. Since it is given that remainder in this condition is 4. Hence, the lease number to be subtracted = (176 – 4)172. (d) |

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