By Ivan Niven, Herbert S. Zuckerman, Hugh L. Montgomery

The 5th version of 1 of the traditional works on quantity thought, written by way of internationally-recognized mathematicians. Chapters are particularly self-contained for higher flexibility. New positive aspects contain increased remedy of the binomial theorem, ideas of numerical calculation and a piece on public key cryptography. includes a good set of difficulties.

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**Example text**

Solution: The answer is 207. Note that digits 4, 6, and 8 cannot appear in the units digit. Hence the sum is at least 40 + 60 + 80 + 1 + 2 + 3 + 5 + 7 + 9 = 207. On the other hand, this value can be obtained with the set {2, 5, 7, 43, 61, 89}. 44. Write 101011(2) in base 10, and write 1211 in base 3. Solution: We have 1010011(2) = 1 · 26 + 0 · 25 + 1 · 24 + 0 · 23 + 0 · 22 + 1 · 2 + 1 = 64 + 16 + 2 + 1 = 83. Dividing by 3 successively, the remainders give the digits of the base-3 representation, beginning with the last.

For each n = 1, 2, 3, . . , 15, there are 100 − 6n suitable values of i (and j), so the number of solutions is 94 + 88 + 82 + · · · + 4 = 784. Euler’s Totient Function We discuss some useful properties of Euler’s totient function ϕ. 31. Let p be a prime, and let a be a positive integer. Then ϕ( pa ) = pa − pa−1 . 32. Let a and b be two relatively prime positive integers. Then ϕ(ab) = ϕ(a)ϕ(b). Proof: : Arrange the integers 1, 2, . . , ab into an a × b array as follows: 1 a+1 .. 2 a+2 .. ··· ··· ..

But n = a0 + a1 b + · · · + ak bk ≤ (b − 1)(1 + b + · · · + bk ) = bk+1 − 1 < bk+1 , a contradiction. If h = k, then a 0 + a 1 b + · · · + a k b k = c0 + c 1 b + · · · + c k b k , and so b | (a0 − c0 ). On the other hand, |a0 − c0 | < b; hence a0 = c0 , Therefore a1 + a2 b + · · · + ak bk−1 = c1 + c2 b + · · · + ck bk−1 . By repeating the above procedure, it follows that a1 = c1 , a2 = c2 , . . , and a k = ck . Relation (∗) is called the base-b representation of n and is denoted by n = ak ak−1 .