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Re: How many integers from 0 to 1000 inclusive, have a remainder of 3 when [#permalink]

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30 Sep 2017, 06:25

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RMD007 wrote:

How many integers from 0 to 1000 inclusive, have a remainder of 3 when divided by 16?

A. 61

B. 62

C. 63

D. 64

E. 65

Source: Self-made.

Combined approach: using arithmetic, find first and last term with remainder 3. Then calculate the number of multiples of 16 in an evenly spaced sequence.

First term: \(\frac{0}{16}\) = 0 + R3, first term is 3

Last term: is 16(n) + 3 \(\leq{1,000}\). Ignore the 3 for a bit.

\(\frac{1,000}{16} = 62.x\)

\((16)(62) = 992\). Now add 3. \((992+3) = 995\) = Last Term

Number of terms: \(\frac{Last-First}{Increment}+ 1\) =

\(\frac{995-3}{16} + 1\) = \(62 + 1 = 63\)

ANSWER C

Shortcut: Above calculation shows 62 multiples of 16 in 1,000. Highest multiple + 3 is not > 1,000. Add one more multiple (because for \(\frac{1,000}{16}\), first multiple of 16 is 16): 0 is a multiple of every integer. 62 + 1 = 63 terms

I think you found an approach with a smart pattern that works! It is a bit similar to mine, above.

I do not think your approach is "brute force" except maybe in one place where, had the upper limit of the range been an ugly number, you might have been calculating for awhile.

I didn't use round multiples of 10, times 16, to "multiply up" to the highest term. Maybe that is faster. It is certainly smart. Next time, if numbers aren't bad, I'll try your way. Just one tiny cautionary thought . . .

If you were to get some heinous number, say 173,619, as the last of the range, you might be multiplying for awhile, even if you used thousands, hundreds, and then tens. In a case such as that, to find multiples, you may want to try the reverse: divide the heinous number by the multiple to get the greatest number that "n" in 16n can be.

In other words: to find the highest multiple of 16 close to but less than 1,000: divide 1000 by 16 to get 62.xx.

It's 62.5, but I didn't calculate that. The second I saw I had a decimal coming, I just noted the .xx. Then I took the integer, 62, and multiplied by 16 to give me that highest value, 992. Added 3. Done on that front (very similar to your conclusion). If this cautionary thought seems dumb, of course, ignore it.