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:
there are total three numbers that satisfy the given conditions :6360,1765 and 7315
Step-by-step explanation:
There are two possible cases for numbers divisible by 5.
They have 5 or 0 in the unit place.
So,
Case 1: The number imagined is ABC5
Now according to the given conditions, if the first digit is erased then we have BC5 and it is a multiple of 9
=>B+C+5=9,18
As B and C are digits they can take a maximum value of 9 and a minimum value of 0
=>B+C=4 or B+C=13
if the second digit is erased then we have AC5 and it is a multiple of 11=>AC5 is divisible by 11
if the third digit is erased then we have AB5 and it is a multiple of 7
AB5 =105,175,245,385,455,525,595,665,735,875,945
ABC5={105-1045},{175-1765},{245-2495|2405},{385-3855},{455-4585},{525-5225},{665-6675},{735-7315},{875-8765},{945-9495|9405}
AC5=145,165,295,205,355,485,525,675,715,865,995,905
Among these AC5 those divisible by 11 are 165,715
Inserting their B respectively:
1765,7315
Case 2: The number imagined is ABC0
If A is removed then we have BC0 which is divisible by 9 => B+C =9
If B is removed we have AC0 which is divisible by 11 => A=C
If C is removed we have AB0 which is divisible by 7 => AB/CB is a multiple of 7
So B=3 and C=6
and since A=C => A=6
so ABC0=6360