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Problems for the
Middle School
Problem Editor : R. ATHMARAMAN
Problems for Solution
The problems in this selection are all woven
around the theme of GCD (‘greatest common
divisor’, also called ‘highest common factor’) and
LCM (‘least common multiple’).
Problem II-1-M.1
Two-digit numbers a and b are chosen (a > b).
Their GCD and LCM are two-digit numbers, and
a/b is not an integer. What could be the value of
a/b?
Problem II-1-M.2
The sum of a list of 123 positive integers is 2013.
Given that the LCM of those integers is 31, find
all possible values of the product of those 123
integers.
Problem II-1-M.3
Let a and b be two positive integers, with a ≤ b,
and let their GCD and LCM be c and d,
respectively. Given that a + b = c + d, show that:
(i) a is a divisor of b; (ii) a3 + b3 = c3 + d3.
Problem II-1-M.4
Let a and b be two positive integers, with a ≤ b,
and let their GCD and LCM be c and d,
respectively. Given that ab = c + d, find all
possible values of a and b.
Problem II-1-M.5
Let a and b be two positive integers, with a ≤ b,
and let their GCD be c. Given that abc = 2012,
find all possible values of a and b.
Problem II-1-M.6
Let a and b be two positive integers, with a ≤ b,
and let their GCD and LCM be c and d,
respectively. Given that d − c = 2013, find all
possible values of a and b.
Solutions of Problems in Issue-I-2
Solution to problem I-M-S.1 Using the digits
0, 1, 2, 3, 4, 5, 6, 7, 8, 9 once each, can you make
a set of numbers which when added and
subtracted in some order yields 100?
If the problem had said only ‘added’ (with
subtraction not allowed) the answer is that this
is not possible! For, the sum
0 + 1 + 2 +···+ 8 + 9 = 45 is a multiple of
9, hence any set of numbers made using these
digits and added together will yield a multiple
of 9. For example, the sum 125 + 37 + 46+
80 + 9 equals 297, which is a multiple of 9.
So an answer of 100 would be impossible to
achieve.
However with subtraction permitted, the task is
possible. Let A represent the part which is ‘added’
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and B the part which is subtracted. Then we want
A − B = 100. For reasons already explained,
A + B ≡ 0 (mod 9); also, A − B ≡ 1 (mod 9).
These two relations yield A ≡ 5 (mod 9) and
B ≡ 4 (mod 9). Our task now is to partition the
digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 into two subsets,
with sums 5 (mod 9) and 4 (mod 9) respectively,
and try to create numbers using the two sets of
digits whose sums differ by 100. One possible
approach is to initially leave out the digits
0, 3, 6, 9 and work only with the digits
1, 2, 4, 5, 7, 8. After some play we find that the
following partition works: {1, 2, 4, 7} and {5, 8};
observe that 1 + 2 + 4 + 7 = 14 ≡ 5 (mod 9) and
5 + 8 = 13 ≡ 4 (mod 9). A convenient possibility
is 72 + 14 − 85 = 1. Now if we can somehow
create 99 using the remaining digits, our task is
done. This is possible: 90+ 6 + 3 = 99. So we have
our answer: 90 + 6 + 3 + 72 + 14 − 85 = 100.
Solution to problem I-M-S.2 To find a formula for
the n-th term of the sequence of natural numbers
from which the multiples of 3 have been deleted:
1, 2, 4, 5, 7, 8, ... .
We make use of the floor function, defined as
follows: [x] = the largest integer not exceeding x.
Example: [2.3] = 2, [10.7] = 10, [−1.7] = −2. Let
f (n) denote the n-th term of the sequence
1, 2, 4, 5, 7, 8, ... . Then the sequence f (n) − n
has the following terms: 0, 0, 1, 1, 2, 2, 3, 3, ... .
The n-th term for this is easy to work out: it is
simply [(n − 1)/2]. Hence f(n) = n + [(n − 1)/2].
Solution to problem I-M-S.3 To find a formula for
the n-th term of the sequence of natural numbers
from which the squares have been deleted:
2, 3, 5, 6, 7, 8, 10, 11, 12, ... .
We again use the floor function. Let g(n) denote
the n-th term of the sequence
2, 3, 5, 6, 7, 8, 10, 11, 12, ... . Then the
sequence g(n) − n has the following terms:
1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, ... .
Note the pattern: two 1s, four 2s, six 3s, eight 4s,
... . The last 1 comes at position 2; the last 2
comes at position 6; the last 3 comes at position
12; ... . It is easy to see that the last k must come
at position k(k + 1). Hence g(n) − n = k precisely
when (k − 1)k < n ≤ k(k + 1). Solving these
inequalities for k we find that
g(n) = n +
[
√4n] + 1
2
.
It turns out that this can be expressed in a much
more pleasing form:
g(n) = n +
n + √n
.
The proof of this surprising equality is left to the
reader.
Solution to problem I-M-S.4 Amar, Akbar and
Antony are three friends. The average age of any
two of them is the age of the third person. Show
that the total of the three friends’ ages is divisible
by 3. By focusing on the age of the oldest among
the three persons, or the youngest among them
(assuming there is an oldest), we easily deduce
that their ages are identical. Hence the sum of the
ages is a multiple of 3.
Solution to problem I-M-S.5 A set of consecutive
natural numbers starting with 1 is written on a sheet
of paper. One of the numbers is erased. The average
of the remaining numbers is 52
9 . What is the number
erased? Let the largest number be n, so the sum of
the numbers is n(n + 1)/2; let the number erased
be x, where 1 ≤ x ≤ n. Then we have the following
equation which we must solve for n and x:
1
2 n(n + 1) − x
n − 1 = 47
9 .
Cross-multiplying and simplifying (we leave the
details to you) we get:
9n2 − 85n + 94 = 18x.
From this we see that 9 | 85n − 94, hence
9 | 4n − 4 = 4(n − 1), hence 9 | n − 1. (Recall that
a | b means: ‘a is a divisor of b’.) Therefore
n ∈ {1, 10, 19, 28, 37, 46,...}.
Next, since 1 ≤ x ≤ n, it follows that
1
2 n(n + 1) − n
n − 1 ≤
47
9 ≤
1
2 n(n + 1) − 1
n − 1 .
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We solve these two inequalities for n. The one on
the left gives:
9n(n − 1)
2 ≤ 47(n − 1), ∴ 9n ≤ 94,
∴ n ≤ 10,
since n is a whole number. The one on the right
gives:
9(n − 1)(n + 2)
2 ≥ 47(n − 1), ∴ 9(n + 2) ≥ 94,
∴ n ≥ 9.
Hence n ∈ {9, 10}. Invoking the earlier condition
we get n = 10, and the number removed is
x = (900 − 850 + 94)/18 = 144/18 = 8.
Solution to problem I-M-S.6 The average of a
certain number of consecutive odd numbers is A. If
the next odd number after the largest one is
included in the list, then the average goes up to B.
What is the value of B − A?
The sum of k consecutive odd numbers starting
with 2n + 1 is (k + n)2 − n2 = k2 + 2nk, hence the
average of these numbers is k + 2n. The average
of k + 1 consecutive odd numbers starting with
2n + 1 is clearly k + 1 + 2n. The difference
between these two is 1. Hence B − A = 1.
Solution to problem I-M-S.7 101 marbles
numbered from 1 to 101 are divided between two
baskets A and B. The marble numbered 40 is in
basket A. This marble is removed from basket A and
put in basket B. The average of the marble numbers
in A increases by 1/4; the average of the marble
numbers in B also increases by 1/4. Find the
number of marbles originally present in basket A.
(1999 Dutch Math Olympiad.)
Let baskets A and B have n marbles and 101 − n
marbles at the start, and let the averages of
baskets A and B be x and y, respectively. Then the
totals of the numbers in the two baskets are,
respectively, nx and (101 − n)y. Since the total
across the two baskets is
1 + 2 + 3 +···+ 101 = 101 × 102/2 = 5151, we
have:
nx + (101 − n)y = 5151. (1)
After the transfer of marble #40 from A to B, the
individual basket totals are nx − 40 and
(101 − n)y + 40, and the new averages are,
respectively:
nx − 40
n − 1 , (101 − n)y + 40
102 − n .
We are told that the new averages exceed the old
ones by 1/4. Hence:
nx − 40
n − 1 − x = 1
4
,
(101 − n)y + 40
102 − n − y = 1
4
.
Hence:
nx − 40 − (n − 1)x = n − 1
4 ,
(101 − n)y + 40 − (102 − n)y = 102 − n
4 .
These yield, on simplification:
x − 40 = n − 1
4 , 40 − y = 102 − n
4 . (2)
We must solve (1) and (2). Substituting from (2)
into (1) we get:
n
n − 1
4 + 40
+ (101 − n)
40 − 102 − n
4
= 5151.
This yields:
101(n + 29)
2 = 5151,
∴ n + 29 = 51 × 2 = 102,
giving n = 73.
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