Question

) Find all 7 digit numbers formed by using only the digits 5 and 7 and divisible by both 5 and 7.​

Answer 1

Step-by-step explanation:

The number has to be of the form abcdef5. Putting a 7 at either of the places a,b,c,d,e,f gives a contribution 0 mod 7. Putting a 5 gives a different contribution mod 7 depending on the place where you put it. A 5 at position a,b,c,d,e,f gives a contribution mod 7 of respectively 5,4,6,2,3,1. Each number of the form abcdef5 can be identified with the subset of {a,b,c,d,e,f} of positions where a 5 is put. For the number to be divisible by 7 we require the sum of contributions mod 7 from the first six digits to be 2, as together with the 5 at the end this will give us 0 mod 7. A quick check now gives the following:

Sum 2: {d}

Sum 9: {f,d,c} {f,e,a} {d,e,b} {e,c} {b,a}

Sum 16: {f,d,e,b,c} {d,e,a,c} {f,b,a,c}

So all the numbers are 7775775, 7755755, 5777555, 7575575, 7757575, 5577775, 5777775, 5755575 and 5557755.

Answer 2

Answer:

I think maybe it's 35 becoz 5and7 can be multiplied and as well be divisible to 35