A,B,C and D are positive integers less than 10.A is 3 times of B.The difference between B and D is 2. C is 1 less than D.Which is the smallest number?
Anonymous:
Actually there are no such integers which satisfy above conditions
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A is 3 times B, thus B is smaller than A
B = A/3
So B can be 1,2 or 3 as A is less than 10.
Difference between B and D is 2
D can be either B+2 or B-2.
If B=1 or 2, D will be B+2 (as B-2 is -1 and 0, but we know that D is a positive integer). C will be D-1 or B+1. So B will be smallest.
If B=3, then D will be either 1(B-2) or 5(B+2).
C is one less than D, thus C is smaller than D.
C is 0 or 4. But all are positive. So C is 4.
Hence B(=3) is the smallest.
Thus B is smallest.
B = A/3
So B can be 1,2 or 3 as A is less than 10.
Difference between B and D is 2
D can be either B+2 or B-2.
If B=1 or 2, D will be B+2 (as B-2 is -1 and 0, but we know that D is a positive integer). C will be D-1 or B+1. So B will be smallest.
If B=3, then D will be either 1(B-2) or 5(B+2).
C is one less than D, thus C is smaller than D.
C is 0 or 4. But all are positive. So C is 4.
Hence B(=3) is the smallest.
Thus B is smallest.
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1 <= A, B, C, D <= 9
A = 3 B => 1 <= B < 3 and 3 <= A <= 9
C = D - 1 => 1 <= C <= 8 and 2 <= D <= 9
| B - D | = 2
B is maximum 3, D is minimum 2 => B - D <= 1, if B > D.
Hence D > B and D - B = 2 => B = D -2 = C - 1
B = A/3, C - 1 and D - 2 . Hence B is the smallest.
A = 3 B => 1 <= B < 3 and 3 <= A <= 9
C = D - 1 => 1 <= C <= 8 and 2 <= D <= 9
| B - D | = 2
B is maximum 3, D is minimum 2 => B - D <= 1, if B > D.
Hence D > B and D - B = 2 => B = D -2 = C - 1
B = A/3, C - 1 and D - 2 . Hence B is the smallest.
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