Transcript pps

Slide 1

Puzzle dtd 12.11.2009 - Little
HARD

http://www.nidokidos.org

Arrange the digits from 1 to 9 to make a 9-digit
number ABCDEFGHI which satisfies the following
conditions:


1)
2)
3)
4)
5)
6)
7)
8)
9.

AB is divisible by 2;
ABC is divisible by 3;
ABCD is divisible by 4;
ABCDE is divisible by 5;
ABCDEF is divisible by 6;
ABCDEFG is divisible by 7;
ABCDEFGH is divisible by 8;
ABCDEFGHI is divisible by
http://www.nidokidos.org

ANSWER :



From condition 4, we know that E equals 5.
From conditions 1, 3, 5 and 7, 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 3 and 7 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 2 and 5, 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 381654729

http://www.nidokidos.org

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~* ~ Rajesh ~*~
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Slide 2

Puzzle dtd 12.11.2009 - Little
HARD

http://www.nidokidos.org

Arrange the digits from 1 to 9 to make a 9-digit
number ABCDEFGHI which satisfies the following
conditions:


1)
2)
3)
4)
5)
6)
7)
8)
9.

AB is divisible by 2;
ABC is divisible by 3;
ABCD is divisible by 4;
ABCDE is divisible by 5;
ABCDEF is divisible by 6;
ABCDEFG is divisible by 7;
ABCDEFGH is divisible by 8;
ABCDEFGHI is divisible by
http://www.nidokidos.org

ANSWER :



From condition 4, we know that E equals 5.
From conditions 1, 3, 5 and 7, 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 3 and 7 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 2 and 5, 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 381654729

http://www.nidokidos.org

Shared By
~* ~ Rajesh ~*~
o.com>



http://www.nidokidos.org


Slide 3

Puzzle dtd 12.11.2009 - Little
HARD

http://www.nidokidos.org

Arrange the digits from 1 to 9 to make a 9-digit
number ABCDEFGHI which satisfies the following
conditions:


1)
2)
3)
4)
5)
6)
7)
8)
9.

AB is divisible by 2;
ABC is divisible by 3;
ABCD is divisible by 4;
ABCDE is divisible by 5;
ABCDEF is divisible by 6;
ABCDEFG is divisible by 7;
ABCDEFGH is divisible by 8;
ABCDEFGHI is divisible by
http://www.nidokidos.org

ANSWER :



From condition 4, we know that E equals 5.
From conditions 1, 3, 5 and 7, 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 3 and 7 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 2 and 5, 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 381654729

http://www.nidokidos.org

Shared By
~* ~ Rajesh ~*~
o.com>



http://www.nidokidos.org


Slide 4

Puzzle dtd 12.11.2009 - Little
HARD

http://www.nidokidos.org

Arrange the digits from 1 to 9 to make a 9-digit
number ABCDEFGHI which satisfies the following
conditions:


1)
2)
3)
4)
5)
6)
7)
8)
9.

AB is divisible by 2;
ABC is divisible by 3;
ABCD is divisible by 4;
ABCDE is divisible by 5;
ABCDEF is divisible by 6;
ABCDEFG is divisible by 7;
ABCDEFGH is divisible by 8;
ABCDEFGHI is divisible by
http://www.nidokidos.org

ANSWER :



From condition 4, we know that E equals 5.
From conditions 1, 3, 5 and 7, 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 3 and 7 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 2 and 5, 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 381654729

http://www.nidokidos.org

Shared By
~* ~ Rajesh ~*~
o.com>



http://www.nidokidos.org