1. Find the least number whose last digit is 7 and which becomes 5 times larger when this last digit is carried to the beginning of the number.
Answers
Answer:
HOPE IT HELPS
Step-by-step explanation:
I’m going to be a bit more simple-minded than some of the other answers, for those who prefer to just try a few possibilities rather than think too hard about the maths. (At least, that’s my excuse!)
Anyway, let’s call the number with the last digit of 7 10n+7 , so n is the number you get by removing the 7. Let’s also define m as the closest power of 10, rounding up from n . (So, for example, if n was 17, m would be 100.) Then, we can write the question as follows:
5(10n+7)=7m+n.
Rearranging that a little gives us
49n+35=7m,
or
7n+5=m.
Essentially what this is telling us is that we need to find a power of 10 that leaves a remainder of five when divided by 7. (Since 7 and 10 are coprime, we’ll find one of those within 7 tries.) Once we find the lowest such power, it’s simple to find n .
Working our way up, we find that (100000−5)/7=14285 , and hence the answer we want is 142857.
So, there you go. Five tries with the calculator got us there. Now I’m off to learn from all the other answers