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How many integers from 1 to 1000, inclusive, have the same remainder when divided by 2, 3, 5, 7?
- a)6
- b)7
- c)8
- d)9
- e)10
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
How many integers from 1 to 1000, inclusive, have the same remainder w...
To find the integers from 1 to 1000, inclusive, that have the same remainder when divided by 2, 3, 5, and 7, we need to find the least common multiple (LCM) of these four numbers. The LCM of 2, 3, 5, and 7 is 210.
Now, let's divide the range 1 to 1000 by 210. We divide 1000 by 210 and get a quotient of 4 and a remainder of 40. This means that the numbers from 1 to 1000 that have a remainder of 40 when divided by 210 are 40, 250, 460, 670, and 880.
Similarly, we divide 1000 by 210 and get a quotient of 4 and a remainder of 0. This means that the numbers from 1 to 1000 that have a remainder of 0 when divided by 210 are 0, 210, 420, 630, and 840.
Therefore, there are a total of 9 integers from 1 to 1000, inclusive, that have the same remainder when divided by 2, 3, 5, and 7. Hence, the correct answer is option D) 9.
Now, let's divide the range 1 to 1000 by 210. We divide 1000 by 210 and get a quotient of 4 and a remainder of 40. This means that the numbers from 1 to 1000 that have a remainder of 40 when divided by 210 are 40, 250, 460, 670, and 880.
Similarly, we divide 1000 by 210 and get a quotient of 4 and a remainder of 0. This means that the numbers from 1 to 1000 that have a remainder of 0 when divided by 210 are 0, 210, 420, 630, and 840.
Therefore, there are a total of 9 integers from 1 to 1000, inclusive, that have the same remainder when divided by 2, 3, 5, and 7. Hence, the correct answer is option D) 9.
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How many integers from 1 to 1000, inclusive, have the same remainder w...
The least common multiple of 2, 3, 5, and 7 is 210. Therefore, if we express a number N as 210q + 1, where q is an integer, we can find a series of numbers that satisfy this equation.
The first number that satisfies this equation is N = 1 when q = 0.
The second number that satisfies this equation is N = 211 when q = 1.
The third number that satisfies this equation is N = 421 when q = 2.
Hence, we observe that these numbers form an arithmetic progression with a common difference of 210.
The number of terms below 1000 that satisfy the equation N = 210q + 1 is 5.
Therefore, the total number of terms in the sequence is 4 + 5 = 9.
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