SW-ARML 3-11-12
1.
How many four-digit integers have the sum of their two leftmost digits equals the sum of
their two rightmost digits?
Suppose the 2-digit sum is d. We must have 1 ≤ d ≤ 18. For d ≤ 9, there are d + 1 choices for
the 2nd pair and one less for the 1st pair (numbers. cannot have leading 0). For d > 9, there
are 19 – d choices for each pair. Hence 1·2 + 2·3 + ... + 9·10 + 92 + 82 + ... + 12 = 615.
2.
For each positive integer k, let Sk denote the increasing arithmetic sequence of integers
whose first term is 1 and whose common difference is k. For example, S3 is the sequence 1,
4, 7, 10, … For how many values of k does Sk contain the term 1001?
Suppose that the nth term of the sequence Sk is 1001. Then 1 + (n – 1)k = 1001, and (n – 1)k
= 1000 = 2353. There are (3 + 1)(3 + 1) = 16 ways to form the order pair (n – 1, k), and each
of these gives a possible value of k. For example, if n – 1 = 8 and k = 125, then the 9th term
of the sequence 1, 126, 251, …. is 1001.
3.
Find the least positive integer such that when its leftmost digit is deleted, the resulting
1
integer is
of the original integer.
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The number can be represented as 10n a + b, where a is the leftmost digit and b is the rest of
10n a  b
 28b  10n a . Since 7 divides 28b, we must have
the numbers. We know that b 
29
n
a = 7, implying 4b = 10 . The smallest n for this to work is n = 2, so b = 25. The number
we are looking for is 725.
4.
How many positive integers have exactly three proper divisors (positive integral divisors
excluding itself), each of which is less than 50?
Suppose n is such an integer. Then one of its factors is 1, so n must be in the form n = pq or
n = p3 for distinct prime numbers p and q.
In the first case, the three proper divisors of n are 1, p and q. Thus, we need to pick two
prime numbers less than 50. There are fifteen of these (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31,
15 
37, 41, 43 and 47) so there are    105 numbers of the first type.
2
In the second case, the three proper divisors of n are 1, p and p2. Thus we need to pick a
prime number whose square is less than 50. There are four of these (2, 3, 5 and 7) and so
four numbers of the second type.
Thus, the total is 109.
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5.
The digits of a positive integer n are four consecutive integers in decreasing order when read
from left to right. What is the sum of the possible remainders when n is divided by 37?
Finding the remainder when divided by 37 is equivalent to computing mod 37.
Our numbers have the form (n  3)(n  2)(n  1)(n) = 1000( n  3)  100( n  2) 10( n 1)  n
 1111n  3210 for 0  n  6. Now, note that 1111  1 (mod 37) and 3210  28 (mod 37), so
1111n + 3210  n + 28 = 28, 29, 30, 31, 32, 33, or 34. The sum of these remainders is 731 =
217.
6.
The equation 2333x–2 + 2111x+2 = 2222x+1 + 1 has three real roots. Given that their sum is
m
,
n
where m and n are relatively prime positive integers, find m + n.
1 3
y  4 y  2 y 2  1 or y3  8 y 2  16 y  4  0 . The
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product of the three roots y1, y2, & y3 is equal to 4. Let the corresponding values of x be
x1, x2, x3. Then, 4  2111x1  2111x2  2111x3  2111( x1  x2  x3 ) , implying that 111(x1, + x2 + x3) = 2, or
2
m
x1  x2  x3 
 . Thus, m + n = 113.
111 n
Let y = 2111x. Then our equation reads
7.
Corresponding terms of two arithmetic progressions are multiplied to give the sequence
1440, 1716, 1848, ... . Find the eighth term.
Let the first sequence be A – D, A, A + D, A + 2D, and the second sequence be
a – d, a, a + d, and a + 2d. Then, (A – D)(a – d) = 1440, Aa = 1716, (A + D)(a + d) = 1848,
so 2Dd = (A – D)(a – d) + (A + D)(a + d) – 2Aa = 1440 + 1848 – 2·1716 = –144, and
Dd = –72. Hence (Ad + Da) = –[(A – D)(a – d) – Aa – Dd] = –[1440 – 1716 +72] = 204,
so 8th term is (A + 6D)(a + 6d) = Aa + 36Dd + 6(Ad + Da) = 1716 – 36·72 + 6·204 = 348.
8.
How many positive integer divisors of 20122012 are divisible by exactly 2012 positive
integers?
The prime factorization of 2012 is 22503. Thus, the prime factorization of 20122012 is
240245032012. A positive factor of 20122012 will be of the form 2a503b and has (a + 1)(b + 1)
divisors, so we need to solve (a + 1)(b + 1) = 2012 = 22503. There are 32 = 6 possibilities:
(a + 1, b + 1) = (2012, 1), (4, 503), (1006, 2), (2, 1006), (503, 4), (1, 2012);
(a, b) = (2011, 0), (3, 502), (1005, 1), (1, 1005), (502, 3), (0, 2011);
and the factors are 22011, 23503502, 21005503, 25031005, 25025033, & 5032011.
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9.
The roots of x4 – x3 – x2 – 1 = 0 are a, b, c, d. Find p(a) + p(b) + p(c) + p(d), where
p(x) = x6 – x5 – x3 – x2 – x.
x4 – x3 – x2 – 1 = (x+1)(x3 – 2x2 + x – 1), so a = –1, b + c + d = 2, bc + cd + db = 1. Hence,
b2 + c2 + d2 = 22 – 2 = 2. We have p(x) = (x3 – 2x2 + x– 1)(x3 + x2 + x + 1) + x2 – x +1. Hence,
p(a) = 3, p(b) = b2 – b + 1, p(c) = c 2– c +1, p(d) = d2 – d + 1, so p(a) + p(b) + p(c) + p(d)
= 6 + (b2 + c2 + d2) – (b + c + d) = 6.
10. Let C be the coefficient of x2 in the expansion of the product
(1 – x)(1 + 2x)(1 – 3x)(1 + 14x)(1 – 15x). Find C.
Let our polynomial be P(x). It is clear that the coefficient of x is –1 + 2 – 3 +  + 14 – 15 =
–8, so P(x) = 1 – 8x + Cx2 + Q(x), where Q(x) is some polynomial divisible by x3.
Then, P(–x) = 1 + 8x + Cx2 + Q(–x), P(x)P(–x) = 1 + (2C – 64)x2 + R(x), where R(x) is some
polynomial divisible by x3. However, we also know that
P(x)P(–x) = (1 – x)(1 + x)(1 + 2x)(1 – 2x)(1 – 3x)(1 + 3x) (1 – 15x)(1 + 15x)
= (1 – x2)(1 – 4x2) (1 – 225x2) = 1 – (1 + 4 +  + 225)x2 + R(x). Therefore,
2C – 64 = –(1 + 4 +  + 225) = –1240, so 2C = –1176, C = –588.
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