## Gayatri Vidya Parishad student Interview Question

Students**Country:**India

**Interview Type:**In-Person

the no. exists I got it. i think you didn't get the question. suppose take an example no. as 128. if we multiply 128 with 2 to 6 numbers i.e.., 128*2 or 128*3 or 128*4 or 128*5 128*6 then we have to get either 281 or 182 or 812 or 218 or 821 means permutations of 128.

Yeah, this is not an easy problem to solve outright by logic. So far all I've determined is that this number would have to have a fair amount of digits and begin with 1. Also, if one such number exists, infinitely many do, as you could add any number of zeros at the end.

Answer is

22/7=3.142857142857142857142857142857

1 × 142,857 = 142,857

2 × 142,857 = 285,714

3 × 142,857 = 428,571

4 × 142,857 = 571,428

5 × 142,857 = 714,285

6 × 142,857 = 857,142

I thought the question was asking for multiplication with ALL numbers 2..6.

Yes, such number exists.

142857 (1428570, 14285700, 142857000) is a good example, anyone can share a way to logically find such numbers (like a rule with pen and paper).

here are some more:

142857

142857 x 2 = 285714

142857 x 3 = 428571

142857 x 4 = 571428

142857 x 5 = 714285

142857 x 6 = 857142

1428570

1428570 x 2 = 2857140

1428570 x 3 = 4285710

1428570 x 4 = 5714280

1428570 x 5 = 7142850

1428570 x 6 = 8571420

1429857

1429857 x 2 = 2859714

1429857 x 3 = 4289571

1429857 x 4 = 5719428

1429857 x 5 = 7149285

1429857 x 6 = 8579142

14285700

14285700 x 2 = 28571400

14285700 x 3 = 42857100

14285700 x 4 = 57142800

14285700 x 5 = 71428500

14285700 x 6 = 85714200

14298570

14298570 x 2 = 28597140

14298570 x 3 = 42895710

14298570 x 4 = 57194280

14298570 x 5 = 71492850

14298570 x 6 = 85791420

14299857

14299857 x 2 = 28599714

14299857 x 3 = 42899571

14299857 x 4 = 57199428

14299857 x 5 = 71499285

14299857 x 6 = 85799142

142857000

142857000 x 2 = 285714000

142857000 x 3 = 428571000

142857000 x 4 = 571428000

142857000 x 5 = 714285000

142857000 x 6 = 857142000

142985700

142985700 x 2 = 285971400

142985700 x 3 = 428957100

142985700 x 4 = 571942800

142985700 x 5 = 714928500

142985700 x 6 = 857914200

142998570

142998570 x 2 = 285997140

142998570 x 3 = 428995710

142998570 x 4 = 571994280

142998570 x 5 = 714992850

142998570 x 6 = 857991420

142999857

142999857 x 2 = 285999714

142999857 x 3 = 428999571

142999857 x 4 = 571999428

142999857 x 5 = 714999285

142999857 x 6 = 857999142

1428570000

1428570000 x 2 = 2857140000

1428570000 x 3 = 4285710000

1428570000 x 4 = 5714280000

1428570000 x 5 = 7142850000

1428570000 x 6 = 8571420000

1429857000

1429857000 x 2 = 2859714000

1429857000 x 3 = 4289571000

1429857000 x 4 = 5719428000

1429857000 x 5 = 7149285000

1429857000 x 6 = 8579142000

1429985700

1429985700 x 2 = 2859971400

1429985700 x 3 = 4289957100

1429985700 x 4 = 5719942800

1429985700 x 5 = 7149928500

1429985700 x 6 = 8579914200

1429998570

1429998570 x 2 = 2859997140

1429998570 x 3 = 4289995710

1429998570 x 4 = 5719994280

1429998570 x 5 = 7149992850

1429998570 x 6 = 8579991420

1429999857

1429999857 x 2 = 2859999714

1429999857 x 3 = 4289999571

1429999857 x 4 = 5719999428

1429999857 x 5 = 7149999285

1429999857 x 6 = 8579999142

Looks like 142xxxx857 is a pattern, where xxxx can be any number of 9 (digit 9).

If I understand the question correctly, then such a number does not exist.

- amustea June 28, 20122*3*4*5*6 = 720, and the number (X*720) will have more digits than the original number (X).