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Secondary 1 | Maths
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Moonkake
Moonkake

Secondary 1 chevron_right Maths chevron_right Singapore

I m stuck,i first got it but i forget what it is!
Please help and tell me where i went wrong. Thanks
Note that all digits 1 to 9 are used and r not repeated.

Date Posted: 1 month ago
Views: 13
J
J
1 month ago
3816547290
Moonkake
Moonkake
1 month ago
Thankyou
CheesePrata
CheesePrata
1 month ago
From condition 5, we know that e equals 5.

From condition 10, we know that j equals 0

From conditions 2, 4, 6 and 8, we know that b, d, f, h are even numbers, therefore a, c, g, i are 1, 3, 7, 9 in some order.

Furthermore, from conditions 4 and 8 we know that cd is divisible by 4 and gh is divisible by 8 (because fgh is divisible by 8 and f is even). Because c and g are odd, d and h must be 2 and 6 in some order.

From conditions 3 and 6, we know that a+b+c, d+e+f, g+h+i are all divisible by 3.

If d=2, then f=8, h=6, b=4. a+4+c is divisible by 3, therefore a and c must be 1 and 7 in some order, g and i must be 3 and 9 in some order. g6 is divisible by 8, therefore g=9. But neither 1472589 nor 7412589 is divisible by 7.

Therefore d=6, and f=4, h=2, b=8. g2 is divisible by 8, therefore g=3 or 7. a+8+c is divisible by 3, therefore one of a and c is chosen from 1 and 7, the other is chosen from 3 and 9.

If g=3, then one of a and c is 9, the other is chosen from 1 and 7. But none of 1896543, 7896543, 9816543 and 9876543 is divisible by 7.

Therefore g=7, one of a and c is 1, the other is chosen from 3 and 9. From 1836547, 1896547, 3816547 and 9816547, only 3816547 is divisible by 7 (the quotient is 545221).

Therefore, the number we are looking for is 3816547290.

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CheesePrata
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25 answers (A Helpful Person)
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