Problems

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Found: 1936
Problem #PRU-5100
 11–14  3.0

Find all possible non-zero digits \(A\) for which the following holds \((AA+AA+1) \times A = AAA\). (Recall \(AA\) means the two-digit number whose first and second digits are \(A\))

Problem #PRU-5101
 11–14  3.0

A square has been divided into \(4\) rectangles and a square. If the rectangle in the bottom left corner has dimensions \(1 \times 4\) and the one in the top right is \(2 \times 5\), what is the area of the small square in the middle?

Problem #PRU-5102
 11–14  3.0

There are \(25\) bugs sitting on the squares of a \(5 \times 5\) board, \(1\) at each square. When I clap my hands, each bug jumps to a square diagonally from where it was before. Show that after I clap my hands, at least \(5\) squares will be empty.

Problem #PRU-5103
 11–14  3.0

In a convex quadrilateral \(ABCD\), all the triangles \(\triangle ABC\), \(\triangle BCD\), \(\triangle CDA\) and \(\triangle DAB\) have equal perimeters. Show that \(ABCD\) is a rectangle.

Problem #PRU-5104
 11–14  2.0

Find the last two digits of the number \[33333333333333333347^4 - 11111111111111111147^4\]

Problem #PRU-5105
 11–14  2.0

Replace all stars with ”+” or ”\(\times\)” signs so the equation holds: \[1*2*3*4*5*6=100\] Extra brackets may be added if necessary. Please write down the expression into the answer box.

Problem #PRU-5106
 11–14  2.0

In how many ways can one change \(\pounds 2\) into coins worth \(50\)p, \(20\)p and \(10\)p? One does not necessarily need to use all available coin types, i.e. having \(5\) coins of \(20\)p and \(10\) coins of \(10\)p is allowed.

Problem #PRU-5107
 11–14  2.0

Consider two congruent triangles \(ABC\) and \(A_1B_1C_1\). We draw a point \(M\) on the side \(BC\) and a point \(M_1\) on the side \(B_1C_1\) such that the ratio of lengths \(BM:MC\) is equal to the ratio of lengths \(B_1M_1:M_1C_1\). Prove that \(AM = A_1M_1\).
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Problem #PRU-5108
 11–14  2.0

We call a median the segment from the vertex of a triangle to the midpoint of the opposite side. Prove that in two congruent triangles, the corresponding medians are of equal length.

Problem #PRU-5109
 11–14  2.0

We call a bisector the segment from the vertex of a triangle to the opposite side which divides in half the angle next to the starting vertex. Prove that in two congruent triangles, the corresponding bisectors are of equal length.