A meeting of Number Family -- II
Number Theory
Level
4
Find the LARGEST 5-digit-integer N so that 2N is also a 5-digit-integer and all digits 0, 1, 2, 3, ..., 9 are contained in both N and 2N
Note: Each number: N and 2N must contain 5 distinct digits and the digits in N must be different from those in 2N.
The answer is 48651.
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1 solution
We have to maximize the digits...
4 has to be first digit of N. Giving us 9 as first digit of 2N ( 2*4+1 carryover)
Second digit of N cannot be 9, Since 9 is already first digit in 2N.
So let 8 be second digit of N. We have 2*8 = 16 +1 carryover = 17 giving 7 as the second digit of 2N.
Now let 6 be the 3rd digit of N. we have 2*6 = 12 + 1 carryover = 13 giving us 3 as the 3rd digit of 2N.
Now let 5 be the 4th digit of N. We have 2*5 = 10 giving us 0 as the 4th digit of 2N.
Finally we have 1 as the fifth digit of N giving us 2 as the last digit of 2N.
so its 48651 and 97302.