Divisibility by 7
Some divisibility criteria are being taught in elementary school, some in middle school. The following is the summary of what a high schooler ought to know.
n mod 7 = sum( d[k] (10^k mod 7), over k )
Quick check in Mathematica shows that 10^k mod 7 is periodic with period 6:
This gives the divisibility by 7 test. The number is divisible by 7 if the cyclic sum of its digits (starting with rightmost one) with {1,3,2,-1,-3,-2} is divisible by 7. The following is a Mathematica code to compute this cyclic sum: Consider an easy example. Let n = 28. Since 8 mod 7 = 1, we have (8 mod 7)*1+2*3 = 7, and, of course, 28 = 7*4. Now take n=2009. We do (9 mod 7)*1 + 0*3+0*2+2*(-1) = 2-2 = 0.
Indeed:When manually applying the divisibility test for large numbers, it might be easier to use the fact that 1000^k mod 7 is (-1)^(k-1). Thus, we might first form an alternating sum of triples of digits, and test the result afterwards. For example, let n = 123,456,789. It is divisible by 7 if 789 - 456+123 = 333+123 = 456 is divisible by 7. To check this, we compute 6*1+5*3+4*2 = 6+15+8=29 = 1 mod 7. Thus the number n is not divisible by 7. Indeed,Applying the divisibility by 7 directly we get
- A number is divisible by 2, 5, or 10 if its last digit is divisible by, correspondingly, 2, 5, or 10.
- A number is divisible by 3, or 9 if the total of its digits is divisible by, correspondingly, 3, or 9.
- A number is divisible by 4 if its last 2 digits are divisible by 4.
- A number is divisible by 8 if its last 3 digits are divisible by 8.
- (a*b) mod n = (a mod n)*(b mod n) mod n
- (a+b) mod n = (a mod n) + (b mod n) mod n
Quick check in Mathematica shows that 10^k mod 7 is periodic with period 6:
In[30]:= Table[Mod[PowerMod[10, k, 7], 7, -3], {k, 0, 12}]
Out[30]= {1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1}
DivisibilityBySevenTest[n_Integer] :=
With[{dig = Mod[Reverse[IntegerDigits[n]],7]},
dig.PadRight[{},Length[dig], {1,3,2,-1,-3,-2}]
]
In[47]:= DivisibilityBySevenTest[28]
Out[47]= 7
In[48]:= DivisibilityBySevenTest[2009]
Out[48]= 0
In[46]:= 2009/7
Out[46]= 287
In[50]:= QuotientRemainder[123456789, 7]
Out[50]= {17636684, 1}
In[53]:= DivisibilityBySevenTest[123456789]
Out[53]= -13
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