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38%
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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.
A. 61
B. 62
C. 63
D. 64
E. 65
Source: Self-made.
30 Sep 2017, 05:39
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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.
A. 61
B. 62
C. 63
D. 64
E. 65
Source: Self-made.
First no that will leave a remainder of 3 when divide by 16, \(a=3\)
Last no that will leave the remainder 3 when divide by 16 \(T_n= 995\)
every no from 3 to 995 will have a difference between them \(d=16\). Hence using AP formula
\(T_n=a+(n-1)d\) or \(995=3+(n-1)*16\)
Thus \(n=63\)
Option C
30 Sep 2017, 06:25
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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.A. 61
B. 62
C. 63
D. 64
E. 65
Source: Self-made.
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
