27*23n+17*102n
27*23=621
17*102=1734
621+1734=2355
2355n
Factors of 2355n
2355n/n=2355
2355/5=471
471/3=157
2355n=n*5*3*157
11 is NOT a factor
First, by induction it is shown that 102n-1 ≡ -1 (mod 11) for all natural numbers n (that is, all odd powers of 10 are one less than a multiple of 11.) i) When n = 1, then we have 102(1)-1 = 10, which is certainly congruent to -1 (mod 11). ii) Assume 102n-1 ≡ -1 (mod 11). Then, 102(n+1)-1 = 102n+2-1 = 102×102n-1 102 ≡ 1 (mod 11) and 102n-1 ≡ -1 (mod 11), so their product, 102×102n-1, is congruent to 1×(-1) = -1 (mod 11). Now that it is established that 10, 1000, 100000, ... are congruent to -1 (mod 11), it is clear that 11, 1001, 100001, ... are divisible by 11. Therefore, all integer multiples of these numbers are also divisible by 11. A palindrome containing an even number of digits may always be written as a sum of multiples of such numbers. The general form of such a palindrome, where all the As are integers between 0 and 9 inclusive, is A0 100 + A1 101 + ... + An 10n + An 10n+1 + ... + A1 102n + A0 102n+1 which may be rewritten as A0 (100 + 102n+1) + A1 (101 + 102n) + ... + An (10n + 10n+1) and again as 100 A0 (1 + 102n+1) + 101 A1 (1 + 102n-1) + ... + 10n An (1 + 101) Each of the factors in parentheses is one more than an odd power of 10, and is hence divisible by 11. Therefore, each term, the product of one such factor with two integers, is divisible by 11. Finally, the sum of terms divisible by 11 is itself divisible by 11. QED
11 + 11 + 88 = 11*(1 + 1 + 8) = 11*(10) = 110
No. It is divisible by 11.No. It is divisible by 11.No. It is divisible by 11.No. It is divisible by 11.
No, it is not, but 121 and 132 are divisible by 11.
Yes. 11 is divisible by 1 and itself.
No. 6,743,021 is not evenly divisible by 11.
Yes. 1,793 is evenly divisible by 11.
yes and if you ever want to know if something is divisible by 3 add up the number like 3 plus 3 (33) is 6 and it is divisible by 3. by the way the answer is 11. 11x3=33
11 is divisible by no whole numbers except 1 and 11.
No. For instance, 101 is not divisible by 11.
if the difference of the sum at the alternate places is divisible by 11 then it is divisible by 11
Divide it by 11. If the answer is a whole number, it's divisible.