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Can anyone help me? (1 Viewer)

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Can anyone help me?
The first digit of a 6-digit number is 1. This digit 1 is now moved from the first digit position to the end, so it becomes the last digit. The new 6-digit number is now 3 times larger than the original number. What are the last three digits of the original number?

So far I have only managed to expand to and I'm really not sure if I've made a mistake or how to keep going

3(1abcde) =abcde1
300000+30000a+3000b+300c+30d+3e = 100000a+10000b+1000c+100d+10e+1
299999 = 70000a+7000b+700c+70d+ 7e
42857 = 10000a+1000b+100c+10d+e

Please help, any advice is needed,
Cheers
 

KittenToes

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The expansion was a good idea, but I think it would have taken approximately forever. I'm a bit prone to too little working, so let me know if you're still confused. (Note that terms f and g are not equivalent to d or e as the multiplication would have affected them.)

The first digit of a 6-digit number is 1. This digit 1 is now moved from the first digit position to the end, so it becomes the last digit. The new 6-digit number is now 3 times larger than the original number. What are the last three digits of the original number?

3(1abcde) = abcde1

e*3 = f1
3 6 9 12 15 18 21 24 27
Therefore e=7

30*d+3*e = ge1
substitute 7 for e
30*d+21=g71
30*d=g50
30 60 90 120 150 180 210 240 270 300
Therefore d=5

Hence abcde1=abc571
 

001kimchii

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Hence abcde1=abc571
I think the question is referring to the FIRST number created, as in "The first digit of a 6-digit number is 1". The one you said is the "new" number created(as it says in the question). I believe the answer should then be 1AB857.
 

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