2's in 22's in November 22

Number Theory Level 2

$\begin{aligned} 1122 &\equiv 0 \text{ (mod }2) \\ 11223344 &\equiv 0 \text{ (mod }2^2) \\ 112233445566 &\equiv 0 \text{ (mod }2^3) \\ 1122334455667788 &\equiv 0 \text{ (mod }2^4) \end{aligned}$

How many of the congruences above are true?

None of the answers. 1 3 4 2

1 solution

Chew-Seong Cheong
Nov 21, 2018

Since 1122 is even, $1122 \equiv 0 \text{ (mod 2)}$ -- True
$11223344 \equiv 11223300 + 44 \equiv 112233 \times 5^2 \times 2^2 + 44 \equiv 44 \equiv 0 \text{ (mod }2^2)$ -- True
Similarly, $112233445566 \equiv 566 \equiv 6 \not \equiv 0 \text{ (mod }2^3)$ -- False
And $1122334455667788 \equiv 7788 \equiv 12 \not \equiv 0 \text{ (mod }2^4)$ -- False

Therefore, $\boxed 2$ equations are true.

2's in 22's in November 22

1 solution

0 pending reports