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When 7179 and 9699 are divided by another natural number N , remainder obtained is same. How many values of N will be ending with one or more than one zeroes?
- a)24
- b)124
- c)18
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
When 7179 and 9699 are divided by another natural number N , remainder...
9699 - 7129 = 2520.
Prime Factors of 2520: 1, 23, 32, 5, 7.
To get a zero at end: (5*2), (5*22), (5*23)-> 3 ways
& in the remaining numbers 1, 32, 7 -> possible combinations are 1,3,9,7, 21, 63 -> 6 ways.
possible ways of N => 6*3 ways= 18 ways
Prime Factors of 2520: 1, 23, 32, 5, 7.
To get a zero at end: (5*2), (5*22), (5*23)-> 3 ways
& in the remaining numbers 1, 32, 7 -> possible combinations are 1,3,9,7, 21, 63 -> 6 ways.
possible ways of N => 6*3 ways= 18 ways
This question is part of UPSC exam. View all Quant courses
Most Upvoted Answer
When 7179 and 9699 are divided by another natural number N , remainder...
1. Calculate the difference: 7129-9699 = 2520.
2. Prime Factors of 2520: 1, 23, 32, 5, 7.
3. To get a zero at end: (5*2), (5*22), (5*23)-> 3 ways
& in the remaining numbers 1, 32, 7 -> possible combinations are 1,3,9,7, 21, 63 -> 6 ways.
So, possible ways of N => 6*3 ways= 18 ways
Community Answer
When 7179 and 9699 are divided by another natural number N , remainder...
Given: Two numbers 7179 and 9699 leave the same remainder when divided by N.
To find: The number of values of N that end with one or more than one zeroes.
Solution:
Let the remainder obtained when 7179 and 9699 are divided by N be x.
Then, 7179 = Nq + x and 9699 = Nr + x for some integers q and r.
Subtracting the above two equations, we get:
9699 - 7179 = N(r - q)
2520 = N(r - q)
Thus, N divides 2520.
Now, we need to find the number of values of N that end with one or more than one zeroes.
To find this, we need to factorize 2520.
2520 = 2^3 × 3^2 × 5 × 7
A natural number ending with one or more than one zeroes must have a factor of 5 and/or 2.
Thus, N can have the following possible values:
1. 2 × 5 = 10
2. 2^2 × 5 = 20
3. 2^3 × 5 = 40
4. 2 × 5^2 = 50
5. 2^2 × 5^2 = 100
6. 2^3 × 5^2 = 200
7. 2^3 × 5^3 = 1000
8. 2^3 × 5 × 7 = 280
9. 2^2 × 5 × 7 = 70
10. 2 × 5 × 7 = 35
11. 5 × 7 = 35
12. 2^2 × 5 × 3 = 60
13. 5 × 3 = 15
14. 2^2 × 5 × 7 × 3 = 420
15. 2^3 × 5 × 7 × 3 = 840
16. 2^3 × 5^2 × 7 × 3 = 4200
17. 2^2 × 5^2 × 7 × 3 = 2100
18. 2 × 5^2 × 7 × 3 = 1050
Out of these 18 values, only 10, 20, 40, 50, 100, 200, 1000, 700, 60, and 420 have one or more than one zeroes at the end.
Therefore, the correct answer is (c) 18.
To find: The number of values of N that end with one or more than one zeroes.
Solution:
Let the remainder obtained when 7179 and 9699 are divided by N be x.
Then, 7179 = Nq + x and 9699 = Nr + x for some integers q and r.
Subtracting the above two equations, we get:
9699 - 7179 = N(r - q)
2520 = N(r - q)
Thus, N divides 2520.
Now, we need to find the number of values of N that end with one or more than one zeroes.
To find this, we need to factorize 2520.
2520 = 2^3 × 3^2 × 5 × 7
A natural number ending with one or more than one zeroes must have a factor of 5 and/or 2.
Thus, N can have the following possible values:
1. 2 × 5 = 10
2. 2^2 × 5 = 20
3. 2^3 × 5 = 40
4. 2 × 5^2 = 50
5. 2^2 × 5^2 = 100
6. 2^3 × 5^2 = 200
7. 2^3 × 5^3 = 1000
8. 2^3 × 5 × 7 = 280
9. 2^2 × 5 × 7 = 70
10. 2 × 5 × 7 = 35
11. 5 × 7 = 35
12. 2^2 × 5 × 3 = 60
13. 5 × 3 = 15
14. 2^2 × 5 × 7 × 3 = 420
15. 2^3 × 5 × 7 × 3 = 840
16. 2^3 × 5^2 × 7 × 3 = 4200
17. 2^2 × 5^2 × 7 × 3 = 2100
18. 2 × 5^2 × 7 × 3 = 1050
Out of these 18 values, only 10, 20, 40, 50, 100, 200, 1000, 700, 60, and 420 have one or more than one zeroes at the end.
Therefore, the correct answer is (c) 18.
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When 7179 and 9699 are divided by another natural number N , remainder obtained is same. How many values of N will be ending with one or more than one zeroes?a)24b)124c)18d)None of theseCorrect answer is option 'C'. Can you explain this answer?
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