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Question Number 190285 by Abdullahrussell last updated on 31/Mar/23

Commented by ARUNG_Brandon_MBU last updated on 31/Mar/23
#include <stdio.h> int main(void) { unsigned int a, b; for (a = 100; a<1000; a++) { for (b = 300; b<1000; b++) { if (1001a+1 == bb) goto loopend; } } loopend: printf ("a = %u, b = %u", a, b); return 0; }
Commented by ARUNG_Brandon_MBU last updated on 31/Mar/23
Output: a = 183, b = 428
Commented by ARUNG_Brandon_MBU last updated on 31/Mar/23
Sorry! plaintext can only be written as comments not answers.
Commented by Rasheed.Sindhi last updated on 01/Apr/23
However you can write your solution as an 'Answer' in the following way: Empty answer +comment consisting plain text.
Commented by ARUNG_Brandon_MBU last updated on 01/Apr/23
�� ThAnKs
	Answered by talminator2856792 last updated on 31/Mar/23

	$b^{2} - 1 = 1001 a$ $(b - 1) (b + 1) = 1001 a$ $\frac{(b - 1) (b + 1)}{1001} = a$ $\frac{(b - 1) (b + 1)}{7 \times 11 \times 13} = a$ $each of 7, 11 and 13 must divide either b - 1 or b + 1$ $now the necessary combinations for multiplying 7, 11, 13 must be checked .$ $we have six cases :$ $one of the cases :$ $b + 1 is divisible by 7 \times 11 and b - 1 divisible by 13$ $this could be expressed as$ $77 x - 2 \equiv 0 (\mod 13)$ $\Rightarrow 77 x \equiv 2 (\mod 13) for some non negative x .$ $77 \equiv 12 (\mod 13)$ $x = 11$ $b + 1 = 847$ $b = 846$ $a = 715$ $after applying this approach to all other cases$ $the following case yields the value for a that satisfies the conditions .$ $b + 1 is divisible by 11 \times 13 and b - 1 divisible by 7$ $143 x - 2 \equiv 0 (\mod 7)$ $\Rightarrow 143 x \equiv 2 (\mod 7)$ $143 \equiv 3 (\mod 7)$ $x = 3$ $b + 1 = 429$ $b = 428$ $a = 183$
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