Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A three-digit number 'n' (less than 500) is taken. A six-digit number is formed by writing the number 'n' as first three digits and the number '2n' as the last three. Which of the following is necessarily a divisor of the number so obtained?

After this there is actually 4 options to choose $6,37,167,501$. The solution specified in my module is goes like this: "If the original number is $ab$c, the new $6-$digit number will be $abc \times 1002$", after this it is pretty straight forward, but I don't understand how they are forming "$abc \times 1002$" as the concatenated number.

share|improve this question

2 Answers 2

up vote 2 down vote accepted

If you start with n=abc [as digits], then shifting it over by three digits with (abc)*1000 would give abc000. If you wanted abcabc, you could then add n, for (abc)*1000+(abc), or (abc) * 1001. But you want the last three digits to be 2n, so it's (abc)*1000+(abc)*2, or abc*1002.

share|improve this answer

HINT $\ $ Expand $\rm\ (a\ x^2+b\ x + c)\ (x^3 + 2)\ $ then specialize $\rm\ x = 10\:.\:$

As above, since radix notation has polynomial form, many results about decimal or radix notation are special cases of results about polynomials. More generally, in number theory there is a strong analogy between many results about numbers and functions (here polynomials).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.