A boy imagined four-digit multiple of 5 with different digits. If the first digit is erased,
the obtained number is multiple of 9. If the second digit of imagined number is
erased, the obtained number is multiple of 11. If the third digit of imagined number is
erased, the obtained number is multiple of 7. How many numbers satisfy given
conditions
Answers
Given : A boy imagined four-digit multiple of 5 with different digits.
If the first digit is erased, the obtained number is multiple of 9.
If the second digit of imagined number is erased, the obtained number is multiple of 11. If the third digit of imagined number is erased, the obtained number is multiple of 7.
To Find : How many numbers satisfy given conditions
Solution:
four-digit multiple of 5 with different digits.
Hence number can be
ABC5 or ABC0 ( as multiple of 5 end with 0)
Taking case 1
ABC0
first digit is erased, the obtained number is multiple of 9
=> B + C + 0 = multiple of 9
can be B + C = 9 only as different digits
If the second digit of imagined number is erased, the obtained number is multiple of 11
AC0 is multiple of 11 not possible as A and C must be same
Taking case 2
ABC5
first digit is erased, the obtained number is multiple of 9
=> B + C + 5 = multiple of 9 = 9 , 18
B + C = 4 or B + C = 13
If the second digit of imagined number is erased, the obtained number is multiple of 11
=> AC5 is multiple of 11
A - C + 5 = 0
=> A = C - 5 => C > 5 as A is 1st digit and can not be zero
so B + C = 13 only left
possible numbers
1765 , 2675 , 3585 , 4495
3585 and 4495 have repeated digits
1765 and 2675 left
Check last condition
If the third digit of imagined number is erased, the obtained number is multiple of 7.
175 = 25 * 7
265 = 37 * 7 + 6
Hence 1765 is the only possible number
only one number , 1765 Satisfy conditions
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Answer:
Step-by-step explanation: