Asked by lony
the digits of a certain number of 3 arithmetical digits are in arithmetical progression.Their sum is 15.And the number is 129 times the first digit.What is the number???
by using the formula..
please...
by using the formula..
please...
Answers
Answered by
Damon
n, n+1, n+2
3n+3=15
100n+10(n+1)+n+2 = 129n
n = 4
check
400 +50 +6 = 516 ? No
try
n, n+2, n+3
3n+5 = 15
n = 10/3 no
n, n+3, n+6
3n+9 = 15
n = 2
200 + 50 + 8 = 258 YES !
so number is
258
3n+3=15
100n+10(n+1)+n+2 = 129n
n = 4
check
400 +50 +6 = 516 ? No
try
n, n+2, n+3
3n+5 = 15
n = 10/3 no
n, n+3, n+6
3n+9 = 15
n = 2
200 + 50 + 8 = 258 YES !
so number is
258
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