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Find the least wose last digit is 7 and which becomes 5 times larger when this last digit is carried to the beginning of the no.
As I understand, we have a number with n digits, were the last is 7:d1 d2 d3 ... d(n-1) dn = d1 d2 d3 ... d(n-1) 7When the last digit is carried of to the beginning, we have7 d1 d2 d3 ... d(n-1)But, when it holds the number becomes 5 times larger. So, we have:7 d1 d2 d3 ... d(n-1) = 5 x (d1 d2 d3 ... d(n-1) 7)Well, note first that 5 x 7 = 35So, we must have 5 x (d1 d2 ... d(n-1) 7 ) ending in 5, which means that d(n-1) = 5 and:7 d1 d2 d3 ... d(n-2) 5 = 5 x (d1 d2 d3 ... d(n-2) 57)Note now that 5 x 57 = 285So, 5 x (d1 d2 d3 ... d(n-2) 57) ends in 85, which means that d(n-2) = 8 and:7 d1 d2 d3 ... d(n-3) 85 = 5 x (d1 d2 d3 ... d(n-3) 857)Now, 5 x 857 = 4285, which means that d(n-3) = 2Continuing on this way, we eventually obtain the number 142857 which, multiplied by 5 becomes 714285. please rate if it helped u
As I understand, we have a number with n digits, were the last is 7:d1 d2 d3 ... d(n-1) dn = d1 d2 d3 ... d(n-1) 7When the last digit is carried of to the beginning, we have7 d1 d2 d3 ... d(n-1)But, when it holds the number becomes 5 times larger. So, we have:7 d1 d2 d3 ... d(n-1) = 5 x (d1 d2 d3 ... d(n-1) 7)Well, note first that 5 x 7 = 35So, we must have 5 x (d1 d2 ... d(n-1) 7 ) ending in 5, which means that d(n-1) = 5 and:7 d1 d2 d3 ... d(n-2) 5 = 5 x (d1 d2 d3 ... d(n-2) 57)Note now that 5 x 57 = 285So, 5 x (d1 d2 d3 ... d(n-2) 57) ends in 85, which means that d(n-2) = 8 and:7 d1 d2 d3 ... d(n-3) 85 = 5 x (d1 d2 d3 ... d(n-3) 857)Now, 5 x 857 = 4285, which means that d(n-3) = 2Continuing on this way, we eventually obtain the number 142857 which, multiplied by 5 becomes 714285.
please rate if it helped u
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