Question
what is the four digit number which satisfies these conditions;
The last digit is twice the first digit.
The second digit is twice the third digit.
The sum of the first and last digits is twice the third digit.
The last digit is twice the first digit.
The second digit is twice the third digit.
The sum of the first and last digits is twice the third digit.
Answers
a b c d
d = 2 a
b = 2 c
a+d = 2 c
----------------------
a = 2 c - d = 2 c - 2 a
so
3 a = 2 c
hmmmm
if c is 1, nope
need common multiple for integer like
3 a = 6
2 c = 6
looks like a = 2 and c = 3
so far:
2 b 3 d
but d = 2 a
2 b 3 4
b = 2 c so
2 6 3 4
2634
d = 2 a
b = 2 c
a+d = 2 c
----------------------
a = 2 c - d = 2 c - 2 a
so
3 a = 2 c
hmmmm
if c is 1, nope
need common multiple for integer like
3 a = 6
2 c = 6
looks like a = 2 and c = 3
so far:
2 b 3 d
but d = 2 a
2 b 3 4
b = 2 c so
2 6 3 4
2634
thank you verymuch. But how 3a=6?
I needed a common multiple of 2 and 3
that is 6
we are not doing fractions here, just integers
that is 6
we are not doing fractions here, just integers
ok thank you. God blss you
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