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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)46
- 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...
When 7179 and 9699 are divided by another natural number N, remainder obtained is same.
Let remainder is R, then (7179 — R) and (9699 — R) are multiples of N and {(9699 — I?) — (7179 — R)} is multiple of N. Then 2520 is multiple of N or the largest value of N is 2520. Total factors of N which are multiples of 10 is 18.
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Most Upvoted Answer
When 7179 and 9699 are divided by another natural number N , remainder...
Given:
- Two natural numbers 7179 and 9699
- When they are divided by another natural number N, the remainder obtained is the same.
To find:
- Number of values of N that will be ending with one or more than one zeroes.
Solution:
Let the common remainder be 'r'.
Then,
7179 = k1N + r
9699 = k2N + r
Subtracting the above equations, we get:
9699 - 7179 = k2N - k1N
2520 = (k2 - k1)N
As N is a natural number, (k2 - k1) must be a factor of 2520.
The factors of 2520 are:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 18, 20, 21, 24, 28, 30, 35, 36, 40, 42, 45, 56, 60, 63, 70, 72, 84, 90, 105, 120, 126, 140, 168, 180, 210, 252, 280, 315, 360, 420, 504, 630, 840, 1260, 2520
Out of these factors, we need to find the values of N that end with one or more than one zeroes.
A natural number N ends with one or more than one zeroes if and only if it is a multiple of 10.
Therefore, we need to find the factors of 2520 that are multiples of 10.
The factors of 2520 that are multiples of 10 are:
10, 20, 30, 40, 60, 70, 80, 90, 120, 140, 210, 252, 280, 420, 630, 1260, 2520
Hence, the number of values of N that will be ending with one or more than one zeroes is 16.
Therefore, the correct option is (c) 46.
- Two natural numbers 7179 and 9699
- When they are divided by another natural number N, the remainder obtained is the same.
To find:
- Number of values of N that will be ending with one or more than one zeroes.
Solution:
Let the common remainder be 'r'.
Then,
7179 = k1N + r
9699 = k2N + r
Subtracting the above equations, we get:
9699 - 7179 = k2N - k1N
2520 = (k2 - k1)N
As N is a natural number, (k2 - k1) must be a factor of 2520.
The factors of 2520 are:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 18, 20, 21, 24, 28, 30, 35, 36, 40, 42, 45, 56, 60, 63, 70, 72, 84, 90, 105, 120, 126, 140, 168, 180, 210, 252, 280, 315, 360, 420, 504, 630, 840, 1260, 2520
Out of these factors, we need to find the values of N that end with one or more than one zeroes.
A natural number N ends with one or more than one zeroes if and only if it is a multiple of 10.
Therefore, we need to find the factors of 2520 that are multiples of 10.
The factors of 2520 that are multiples of 10 are:
10, 20, 30, 40, 60, 70, 80, 90, 120, 140, 210, 252, 280, 420, 630, 1260, 2520
Hence, the number of values of N that will be ending with one or more than one zeroes is 16.
Therefore, the correct option is (c) 46.
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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)46d)None of theseCorrect answer is option 'C'. Can you explain this answer?
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