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find the smallest no. with 7 in the unit's place such that when the no. is multiplied by 5 ,, 7 is carried to the first place

e.g. if the no. is ab7 ,after multiplying by 5, the no. should be 7ab

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Converted this to a mathematical equation

(10x + 7)*5 = 7*(10^y) + x where x and y are positive integers and y is the no. of digits - 1

simplify this to

x = (7*(10^y) - 35) / 49

Try different values of y, starting from 1 and check what is the lowest value of y for which you get an +ve integer value for x

y x 1 0.7 2 13.6 3 142.1 4 1427.9 5 14285

Therefore the no. is 142857 !

Edited by kya_kay
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Though I am late ....but I find an opportunity to introduce a simpler solution

Put the number as under

abcdef7 * 5 = 7abcdef

so go on multiplying the digits with 5 like this: 5 * 7 = 35

Take unitplce digit as 5 and caryover as 3, now multiply this unitplace digit 5 with 5 and add carryover 3 to it, i.e. 5 * 5 = 25 and 25 + 3 = 28.

Now put 8 of 28 at tenth place an take carryover as 2.

Now multiply 8 (Tenth place digit) with 5 and add carryover 2 to it.

So 5 * 8 = 40, and 40 + 2 = 42. Now put 2 from 42 at 100th place, and carryover as 4. Multiply 2 with 5 and add 4, = 14, put 4 at 1000th place and carry over 1. Multiply 4 with 5 and add 1 = 21.

Put 1 at 10000th place and carryover 2. Multiply 1 with 5 and add 2 to it, = 7.

The number we got is 714285.

This must be the result of muliplying 14285 with 5.

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