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Stanford Math Circle: Sunday, April 11, 2010
Hints, Answers, and Solutions
Warm-up Problems. Choose a few of these problems to work on as you get settled
in today. Once you’ve thought about a problem on your own, talk to someone sitting
near you about your ideas. More challenging problems are marked with (?).
1. Cards with the numbers 7, 8, 9, 4, 5, 6, 1, 2, 3 are laid in a row in the indicated
order. It is permitted to choose several consecutive cards and rearrange them
in the reverse order. Is it possible to obtain the arrangement 1, 2, 3, 4, 5, 6, 7,
8, 9 after three such operations?
2. Consider the following triangle:
1
1 2
1 3 6
. . . .
1
1 1
3 2 1
7 6 3 1
. . . . .
Each number is the sum of the three numbers of the previous row: the number
immediately above it and the numbers immediately to the right and left of that
one. If no number appears in one ore more of these locations, the number zero
is used. Prove that every row, beginning with the third row, contains at least
one even number.
Solution. Beginning with the third row, write the first four numbers of each
row, but putting in place of an even number the letter e and in place of an odd
number the letter θ. Note that the fifth row coincides with the first row. The
evenness or oddness of the first four numbers of every row of the triangle depends
only on the evenness or oddness of the first four numbers in the preceding row.
Thus, any row will periodically duplicate itself in the preceding row. Since an
even number occurs in each of the first four rows, an even number must occur
in every subsequent row.
3. Can you find a way to position 100 lines in the plane to obtain exactly 100
points of intersection? How about 29 lines? 21 lines? 19 lines? 15 lines?
4. How many odd numbers with third digit 5 are there between 20000 and 69999
(inclusive)?
Solution.
5 · 10 · 1 · 10 · 5 = 2500.
5. How many six-digit numbers have at least one even digit?
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Solution. Instead of counting the numbers with at least one even digit, let’s
count the number of six-digit numbers that do not have at least one even digit.
These are exactly the six-digit numbers whose digits are all odd. There are
5 · 5 · 5 · 5 · 5 · 5 = 56 = 15625
of these. Since there are a total of 900,000 six-digit numbers, the number of
six-digit numbers with at least one even digit is
900, 000 − 15625 = 884, 375.
6. The sum of the digits is calculated for the number 2100 , and then the sum of
the digits of the resulting number is calculated. This process continues until a
single digit is left. Which digit is this?
Hint. The final digit is 7. Use remainders modulo 9.
7. Prove that if you reverse the order of the digits in any natural number and
subtract the result from the initial number, then the difference is divisible by 9.
Hint. The original number and the reversed number have equal sums of digits,
and therefore, equal remainders when divided by 9. (Any natural number and
the sum of its digits have the same remainder upon division by 9).
8. An integer is reduced to 1/9 of its value when a certain one of its digits is
deleted, and the resulting number is again divisible by 9.
(a) Prove that division of this resulting integer by 9 results in deleting an
additional digit.
Hint. Prove that the number can be reduced to 1/9 of its value only be
deleting the digit 0, which must be in the second position.
(b) Find all integers satisfying the conditions of the problem.
Answer. 10125; 2025; 30,375; 405; 50625; 6075; 70875. Of course, an
arbitrary number of zeros may be added to each of these numbers.
√
9. Find all integers n which are divisible by all integers not exceeding n.
Answer. Only the following integers satisfy the conditions of the problem:
24, 12, 8, 6, 4, 3, 2.
Hint: prove that all such numbers must be less than 72 = 49.
10. (?) Suppose that the lengths of the legs of a right triangle are perfect squares
(i.e. expressible as the squares of integers). Prove that the hypotenuse cannot
be an integer.
Hint. Show that if x4 +y 4 = z 2 , then there must exist positive integers x1 , y1 , z1 ,
where z1 < z, such that x41 + y14 = z12 . Then prove that there must exist positive
2
integers xN , yn such that x2N + yN
= 1, which is impossible.
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11. (?) Prove that if p is a prime number greater than 3, then the numerator of the
1 1
1
(reduced) fraction 1 + + + · · · +
is divisible by p2 .
2 3
p−1
Hint. The denominator of the sum is (p − 1)!. The numerator of the sum is
the sum of all possible products of the first (p − 1) integers taken (p − 2) at a
time.
12. (a) (?) Ten primes, each less than 3,000, form an arithmetic progression. Find
these prime numbers.
Answer. 199, 409, 619, 829, 1039, 1249, 1459, 1669, 1879, 2089. Hint:
prove that the common difference of the progression must be divisible by
2 · 3 · 5 · 7 = 210.
(b) (?) Can you find eleven primes, all less than 20,000, which form an arithmetic progression? Why or why not?
Answer. No such progression exists. Hint: prove that if the first number
of the progression is not 11, then the common difference must be divisible
by 2·3·5·7·11 = 2310. If the first number is 11, then the common difference
is divisible by 210.
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