Kenneth S. answered • 10/12/17
Tutor
4.8
(62)
Expert Help in Algebra/Trig/(Pre)calculus to Guarantee Success in 2018
every integer is a multiple of 1; 1 should not really be in the question.
not multiple of 2; nor of 3 (because sum of digits is not multiple of 3); not multiple of 4 since not even multiple of 2; not multiple of 5 or 6 or 7 or 8 or 9 either. Must be prime!
Do you know what the criteria for being a multiple of 5 are?
Once it's found to be an odd number, there's no need to ask whether it's a multiple of 4 or 8; similarly once not a multiple of 3, don't bother with testing for divisibility by 32.
Seven is the only tricky one--just do long division to answer that question.
Mark M.
For 7, double last digit, subtract from original, if result is divisible by 7 so is the original. Process can be repeated.
Report
10/12/17
Arthur D.
tutor
Mark, subtract ones digit, divide by 10, then subtract twice original ones digit.
574
574-4=570
570/10=57
57-8=49
686
686-6=680
680/10=68
68-12=56
Report
10/12/17
Kenneth S.
Gentlemen,in my teaching days I decided not to teach the divisibility by 7 rule because I judged it too easy to misremember, over time. So I just do long division and determine if there's no remainder. I do not recall ever seeing the rule in any high school textbook.
Report
10/12/17
Arthur D.
10/12/17