Integer Removal and Sum
Remove the integers which are congruent to 3 (mod 7) from the following list of five integers, and sum the integers that remain. $$
85 \qquad 49,\!479 \qquad -67 \qquad 12,\!000,\!003 \qquad -3
$$
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
Recall that $ a\equiv 3 \pmod{7} $ if and only if $ a-3 $ is divisible by 7. Subtracting 3 from every element in the list gives $$
82 \qquad 49,\!476 \qquad -70 \qquad 12,\!000,\!000 \qquad -6
$$By dividing, we can see that 82 and $ -6 $ are not divisible by 7, whereas $ -70 $ and $ 49,\!476 $ are divisible by 7. To see that $ 12,\!000,\!000 $ is not divisible by 7, note that its prime factorization is $ (12)(10^6)=(2^2\cdot 3)(2\cdot 5)^6 = 2^8\cdot 3\cdot 5^6 $. So, after striking off the numbers which are congruent to 3 (mod 7), the original list becomes $$
85 \qquad \cancel{49,\!479} \qquad \cancel{-67} \qquad 12,\!000,\!003 \qquad -3
$$The sum of the remaining integers is $ \boxed{12,\!000,\!085} $.
Freeflows