The last digit of a six-digit number is 2. If the 2 is moved to the start of the number, the new six-digit number is only a third of the original number. Find the original number.
Please help, I cant get my head around this.
the new number is 2abcde
so 3 times the new number equals the old number
3e must have a ones digit of 2. So it could be equal to 2, 12, 22, 32, etc
But we know it is a digit, so it is less than 10, and since 3*9 = 27, it must equal 2, 12, or 22. And since 12 is the only one of those which is a multiple of 3, it must be 12. then 12/3=4. so e =4
we can't have a ones digit of 12, so we carry the one.
now 3d+1 returns a ones digit of 4. so options are 4, 14, and 24 (because d cannot be greater than 9, so 3d+1 cannot be greater than 28) subtracting one from each of these we get 3d = 3, 13, or 23. and clearly 3 is the only option returning an integer. so 3d=3. thus d = 1
now we know that 3c has a ones digit of 1. so 3c equals 1, 11, or 21 (again, must be less than 27, because c < 9). 21 is the only multiple of 3. so 3c = 21. therefore c = 7
21 is not a digit, so we carry the 2.
now we have that 3b+2 must have a ones digit of 7. so 7, 17, or 27. Subtracting 2 we get that 3b must equal 5, 15, or 25. 15 is the only multiple of 3. so 3b=15. therefore b = 5
17 is not a digit so carry the one.
Now we get that 3a+1 has a ones digit of 5. so 5, 15, or 25. Subtracting one we get that 3a must be 4, 14, or 24. 24 is the only multiple of 3. so 3a=24. therefore a = 8
25 is not a digit, so carry the 2.
Now looking back at the original, equation we see that this becomes
Therefore the number is 857142, and the new number is 285714.
Hello, not happy jan!
Lucky for both of us, I'm familiar with this type of problem . . .
We have: .The last digit of a six-digit number is 2. If the 2 is moved to the front,
the new six-digit number is only a third of the original number.
Find the original number.
And we have a multiplication problem . . .
In column-6, ("carry 1" to column-5)
In column-4, ("carry 2" to column-3)
In column-3, ("carry 1" to column -2)
. . ta-DAA!
Finally took a look at your solution, angel.white . . .
Excellent job . . . nice explanations!