ABACUS International Math Competition
for
5th and 6th graders
April, 1998
B.57. Three travelers met at an intersection.
One of them had 3 loaves of bread, another had 5 loaves of bread, but the
third one had no food. So they divided the bread equally among the three
of them. The third traveler gave the two other travelers $8 for the bread
he received. How should the $8 be fairly distributed between these two travelers?
B.58. There are 5 lamps in a room. Every
one of them can be turned on individually. How many different ways can you
have at least one lamp on?
B.59. How many positive whole numbers with
at most three digits are there in which the sum of the digits is an odd
number, and in the number one greater than the original number the sum of
the digits is an odd number, too?
B.60. Three defendants stand in front of
a judge. Every one of them answers 3 questions, partially with a lie (each
person lies at least once and at most twice):
The answers of A: "It was B." "I have never been on the
scene." "I am innocent."
The answers of B: "C is innocent." "I did not do it."
"All what A says is a lie."
The answers of C: "It wasn't me." "A is lying when he
says that he has never been on the scene." "B is lying when he
says that all what A says is a lie."
Find out who is guilty.
B.61. Four girls participate in a running
race. After the race they were asked what place they finished on.
Anna: "I was neither first nor last."
Bella: "I was not the first at the finish line."
Cecilia: "I was first."
Dory: "I was last."
Somebody, who saw the race, said: "Three of the four answers are
true, and one is false."
Who was the first, and who made the false statement?
B.62. The oldest of three brothers is 14
years older than the youngest. The one in the middle is 4 years younger
than the oldest one. Everybody's age is a prim number. How old are they?
B.63. 1, 9, 9, 8, 7, 3, 7, 5, 2, 7, 1,
5, ...
We created the above sequence by always adding the last four elements
of the sequence and writing down the last digit of this sum as the next
element of the sequence. Can you find 1, 2, 3, 4 (these four numbers in
this order) somewhere in this sequence?
B.64. Can you cut up an equilateral triangle
into 1998 equilateral triangles with no left-over. If yes, how?
by Bognár Ferencné, Hungary
Please, send your solutions to:
Solutions of last year's
problems