WitrynaIt is divisible by the value of its φ function, which returns 48 in its case, and there are 21 solutions to the equation φ(x) = 144. This is more than any integer below it, which makes it a highly totient number. As a square number in decimal notation, 144 = 12 × 12, and if each number is reversed the equation still holds: 21 × 21 = 441. Witryna20 gru 2024 · Fermat's Little theorem states that for every number a not divisible by a …

3.5: The Division Algorithm and Congruence - Mathematics …

Witryna12 sty 2024 · 33)Data Sufficiency ->If p, q, and r are positive integers, where p is an odd number and r = p^2 + q^3 + 4. Is q^3 divisible by 8? (1) r = 18k -5 where k is a positive integer (2) When (r-p+13) is divided by 2, it leaves a remainder. Witryna13 sie 2024 · You are right about the possibility of 4 integers being 0,1,2,3, in which case r = 0. The question statement asks us about r being divisible by 3 i.e. is the remainder 0 when r is divided by 3. If you divide 0 by 3, the remainder is 0. Thus, even if r = 0, we … licensed smog check repair station

144 (number) - Wikipedia

Witryna4 sie 2016 · We can prove this by contradiction. Assume that $4$ divides $n^2 -3$ $$4 n^2-3,$$ then $$4m=n^2-3,$$ where m is some integer. Now let's break the this into two cases. Witryna5 kwi 2015 · If you don't mind going to the next highest integer when there is a tie (e.g. closest integer to 6 that is a multiple of 4 becomes 8 and not 4), then you can use this simple formula. No comparisons are needed. int c = (a + b/2)/b * b; // closest integer to `a` that is divisible by `b` Witryna24 kwi 2024 · I know that for a binary number to be divisible by 3 the sum of 1s in even bits mines the sum of 1s in odd bits must be divisible by 3. After some googling, I found that you can do the dfa of it then convert it to cfg and you can get: \begin{align*} S &\to 0S \mid 1A \mid \epsilon \\ A &\to 0B \mid 1S \\ B &\to 1B \mid 0A \end{align*} licensed small dog breeders in iowa