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Fun Problems
Digital Problems for the Digital Age
Consider all three digit numbers with the
property that the first digit equals the sum of the
second and third digits. Examples of such
numbers are 413, 615 and 404. We call this
property ♥. Let X be the sum of all three digit
numbers that have property ♥.
Next, consider all four digit numbers with the
property that the sum of the first two digits equals
the sum of the last two digits. Examples of such
numbers are 4123, 6372 and 4013. We call this
property ♣. Let Y be the sum of all four digit
numbers with property ♣.
Problem:
Show that both X and Y are divisible by 11.
Note that the problem does not ask for the actual
values of X and Y ; it only asks you to show that
they are multiples of 11. Could there be a way of
proving this without actually computing X and
Y ? We shall show that there is such a way. First,
some notation.
Notation 1: AB denotes the two digit number
with tens digit A and units digit B; ABC denotes
the three digit number with hundreds digitA, tens
digit B and units digit C; ABCD denotes the four
digit number with thousands digit A, hundreds
digit B, tens digit C and units digit D; and so on.
We use the bar notation to avoid confusion, for
example, between the two digit number AB and
the product AB which means A × B.
Notation 2: Given a number with two or more
digits, by its ‘TU portion’ we mean the number
formed by its last two digits. (‘TU’ stands for
‘tens-units’.) For example, the TU portion of 132
is 32, and the TU portion of 1234 is 34.
Notation 3: Given a number with three or more
digits, by its ‘H portion’ we mean its hundreds
digit.
Showing that X is divisible by 11. A three digit
number ABC has property ♥ if A = B + C.
Observe that if ABC has property ♥, so does
ACB. If B = C then these two numbers are the
same. In this case ACB has the form ABB.
Now observe that BB = 11B is a multiple of 11;
so too is BC + CB = 11(B + C). Hence:
• The sum of the TU portions of ABC and ACB
is a multiple of 11.
• The TU portion of ABB is a multiple of 11.
It follows that for each fixed value of A, the sum of
the TU portions of the numbers ABC having
property ♥ is a multiple of 11.
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Now we shall show that the sum of the H portions
of the numbers having property ♥ is a multiple of
11. To show this we adopt a different strategy.
With A = 1 there are two numbers with property
♥ (101 and 110). With A = 2 there are three such
numbers (202, 211 and 220). With A = 3 there are
four such numbers, with A = 4 there are five such
numbers, ... , and with A = 9 there are ten such
numbers. It follows that the sum of the H portions
of the three digit numbers having property ♥ is
(1 × 2) + (2 × 3) + (3 × 4) + (4 × 5)
+ (5 × 6) + (6 × 7) + (7 × 8) + (8 × 9)
+ (9 × 10) = 330,
which is a multiple of 11.
Since the sum of the H portions of all the numbers
with property ♥ is a multiple of 11, and so is the
sum of the TU portions, it follows that X must be a
multiple of 11.
Showing that Y is divisible by 11. We shall use
the same strategy. If ABCD is a number with
property ♣ then A + B = C + D; hence ABDC
too has the property. Since CC = 11C and
CD + DC = 11(C + D) are multiples of 11, it
follows that for each fixed (A, B) pair, the sum of
the TU portions of the numbers ABCD with
property ♣ is a multiple of 11.
Now we focus on the front two digits.
Suppose that ABCD has property ♣, and B is
non-zero. Then BACD too is a four digit number
with property ♣. The sum of the numbers
associated with the front two digits is
AB + BA = 11(A + B), which is a multiple of 11.
What if B = 0? Then the number has the form
A0CD, with A = C + D. This number can be
matched with the three digit number ACD which
has property ♥. We have already shown (in the
above section) that the sum of the A-values of all
such numbers ACD is a multiple of 11. This
proof implies that the sum of the A-values of all
numbers A0CD with property ♣ is a multiple
of 11.
Thus Y is a sum of various multiples of 11, and
hence is a multiple of 11.
It is worth reflecting on the solution strategies
used. We did not at any stage attempt to compute
the actual sum of all the numbers. Instead we
grouped them in a way that would make the
divisibility property perfectly visible.
Problems for Solution
Problem II-1-F.1
Solve the following cryptarithm:
EAT + T H AT = AP P LE.
Problem II-1-F.2
Solve the following cryptarithm:
EART H + MOON = SYST EM.
Problem II-1-F.3
Given that IV × V I = SIX, and SIX is not a
multiple of 10, find the value of IV + V I + SIX.
Problem II-1-F.4
Explain why the following numbers are all perfect
squares:
1, 121, 12321, 1234321,
123454321, 12345654321, ,... .
Problem II-1-F.5
Explain why the following numbers are all perfect
squares:
1089, 110889, 11108889,
1111088889, 111110888889, ... .
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Solutions of Problems from Issue-I-2
Problem I-2-F.1
Show that in a magic triangle, the difference
between the number at a vertex and the number
at the middle of the opposite side is the same for
all three vertices.
We must prove that u − x = v − y = w − z. We
know that u + z + v = v + x + w = w + y + u.
From the equalities we get: u + z − x − w = 0,
hence u − x = w − z. In the same way we get
w − z = v − y.
Hence proved.
u
v w x
z y
Problem I-2-F.2
Explore the analogous problem in which the digits
1, 2, 3, 4, 5, 6, 7, 8, 9 are placed along the sides of
a triangle, one at each vertex and two on the
interiors of each side, so that the sum of the
numbers on each side is the same.
Let the configuration be as shown in the figure,
with the numbers x, y,z at the corners of the
triangle, and the numbers a, b, c, d, e, f on the
interiors of the sides. Then, by requirement, the
sums x + y + e + f , y + z + a + b and
z + x + c + d are all equal to some constant s,
say. Let C = x + y + z be the sum of the corner
numbers, and let M = a + b + c + d + e + f be
the sum of the ‘middle’ numbers.
x
y z
f
e
d
c
a b
By addition we get 3s = 2C + M. Also
C + M = 45; hence C = 3s − 45, and we see that
C is a multiple of 3; so is M. Next, the least
possible value of C is 1 + 2 + 3, and the largest
possible value is 7 + 8 + 9. So 6 ≤ C ≤ 24.
Hence 51 ≤ 3s ≤ 69, leading to 17 ≤ s ≤ 23.
Each s-value between 17 and 23 (and hence each
C-value between 6 and 24 which is a multiple of
3) can be ‘realized’ by a suitable magic triangle.
Two such possibilities are shown below.
Problem I-2-F.3
Show that the cryptarithm
AT + RIGH T = ANGLE has no solutions.
Since the hundreds digits of RIGH T and
ANGLE are the same, we infer that the addition
of AT to RIGH T has only affected the tens and
units digits, with no ‘carry’ to the hundreds digit.
Hence the leading two digits must stay
unaffected; we must have AN = RI . This
violates a basic rule concerning cryptarithms:
that different letters cannot represent the same
digit. Therefore the problem has no solution.
52 At Right Angles | Vol. 1, No. 3, March 2013
Solutions of Problems from Issue-I-2
Problem I-2-F.1
Show that in a magic triangle, the difference
between the number at a vertex and the number
at the middle of the opposite side is the same for
all three vertices.
We must prove that u − x = v − y = w − z. We
know that u + z + v = v + x + w = w + y + u.
From the equalities we get: u + z − x − w = 0,
hence u − x = w − z. In the same way we get
w − z = v − y.
Hence proved.
u
v w x
z y
Problem I-2-F.2
Explore the analogous problem in which the digits
1, 2, 3, 4, 5, 6, 7, 8, 9 are placed along the sides of
a triangle, one at each vertex and two on the
interiors of each side, so that the sum of the
numbers on each side is the same.
Let the configuration be as shown in the figure,
with the numbers x, y,z at the corners of the
triangle, and the numbers a, b, c, d, e, f on the
interiors of the sides. Then, by requirement, the
sums x + y + e + f , y + z + a + b and
z + x + c + d are all equal to some constant s,
say. Let C = x + y + z be the sum of the corner
numbers, and let M = a + b + c + d + e + f be
the sum of the ‘middle’ numbers.
x
y z
f
e
d
c
a b
By addition we get 3s = 2C + M. Also
C + M = 45; hence C = 3s − 45, and we see that
C is a multiple of 3; so is M. Next, the least
possible value of C is 1 + 2 + 3, and the largest
possible value is 7 + 8 + 9. So 6 ≤ C ≤ 24.
Hence 51 ≤ 3s ≤ 69, leading to 17 ≤ s ≤ 23.
Each s-value between 17 and 23 (and hence each
C-value between 6 and 24 which is a multiple of
3) can be ‘realized’ by a suitable magic triangle.
Two such possibilities are shown below.
Problem I-2-F.3
Show that the cryptarithm
AT + RIGH T = ANGLE has no solutions.
Since the hundreds digits of RIGH T and
ANGLE are the same, we infer that the addition
of AT to RIGH T has only affected the tens and
units digits, with no ‘carry’ to the hundreds digit.
Hence the leading two digits must stay
unaffected; we must have AN = RI . This
violates a basic rule concerning cryptarithms:
that different letters cannot represent the same
digit. Therefore the problem has no solution.
52 At Right Angles | Vol. 1, No. 3, March 2013
s = 17, C = 6
1
2 3
9
5
6
7
4 8
s = 23, C = 24
7
8 9
2
6
3
4
1 5
Solutions of Problems from Issue-I-2
Problem I-2-F.1
Show that in a magic triangle, the difference
between the number at a vertex and the number
at the middle of the opposite side is the same for
all three vertices.
We must prove that u − x = v − y = w − z. We
know that u + z + v = v + x + w = w + y + u.
From the equalities we get: u + z − x − w = 0,
hence u − x = w − z. In the same way we get
w − z = v − y.
Hence proved.
u
v w x
z y
Problem I-2-F.2
Explore the analogous problem in which the digits
1, 2, 3, 4, 5, 6, 7, 8, 9 are placed along the sides of
a triangle, one at each vertex and two on the
interiors of each side, so that the sum of the
numbers on each side is the same.
Let the configuration be as shown in the figure,
with the numbers x, y,z at the corners of the
triangle, and the numbers a, b, c, d, e, f on the
interiors of the sides. Then, by requirement, the
sums x + y + e + f , y + z + a + b and
z + x + c + d are all equal to some constant s,
say. Let C = x + y + z be the sum of the corner
numbers, and let M = a + b + c + d + e + f be
the sum of the ‘middle’ numbers.
x
y z
f
e
d
c
a b
By addition we get 3s = 2C + M. Also
C + M = 45; hence C = 3s − 45, and we see that
C is a multiple of 3; so is M. Next, the least
possible value of C is 1 + 2 + 3, and the largest
possible value is 7 + 8 + 9. So 6 ≤ C ≤ 24.
Hence 51 ≤ 3s ≤ 69, leading to 17 ≤ s ≤ 23.
Each s-value between 17 and 23 (and hence each
C-value between 6 and 24 which is a multiple of
3) can be ‘realized’ by a suitable magic triangle.
Two such possibilities are shown below.
Problem I-2-F.3
Show that the cryptarithm
AT + RIGH T = ANGLE has no solutions.
Since the hundreds digits of RIGH T and
ANGLE are the same, we infer that the addition
of AT to RIGH T has only affected the tens and
units digits, with no ‘carry’ to the hundreds digit.
Hence the leading two digits must stay
unaffected; we must have AN = RI . This
violates a basic rule concerning cryptarithms:
that different letters cannot represent the same
digit. Therefore the problem has no solution.
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