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How many 7-digit numbers exist which are divisible by 9 and whose last but one digit is 5?
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- ABCDE5F is divisible by 9.
A+B+C+D+E+F +5 is divisible by 9.
Hence ABCDEF is a 6 digit number of form (9k + 4) where A>1.
Since we have 9*10^5 such numbers and (1/9)th of them are of form (9k + 4)
So 10^5 numbers ( 7 digit ) are divisible by 9 and whose last but one digit is 5 . - 13 years agoHelpfull: Yes(3) No(0)
- largest of such no.s is 9999954 and no.of nines in this are 1111106 but among these nines those divide the no. starting from zero there are 111106 such nines. so the answer is 1111106-111106= 1000000
- 13 years agoHelpfull: Yes(0) No(3)
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