• # question_answer The sum of two numbers is 462 and their highest common factor is 22. What is the minimum number of pair that satisfy these conditions? A)  5                                B)  6            C)  7                                D)  8

Let the required number be 22a and 22b. Then, $22a+22b=462\,$ $22(a+b)=462$ $a+b=21.$ Now, Co - Primes with sum 21 are $(1,\,\,20),\,\,(2,\,\,19),\,\,(4,17),\,\,(5,16),\,\,(8,\,\,13)$and $(10,\,\,11)$ $\therefore$ Required numbers are $(22\times 1,\,\,22\times 20),$$(22\times 2,\,\,22\times 19),$$(22\times 4,\,\,22\times 17),$$(22\times 5,\,\,22\times 16),$$(22\times 8,\,\,22\times 13),$and $(22\times 10,\,\,22\times 11).$ Clearly, the number of such pairs is 6.