This is the third book published by Récréomath.

Mathematical 
Amusements

Par Charles-É. Jean

This edition includes 200 problems and their solution. The French version is published.

Solutions 1 to 50

Solutions 51 to 100

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Solutions 101 to 150
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Solution 101. The square root of 182 is 13,49. One side measures 13 meters and the other 14 meters.

Solution 102. Here is a figure :

Solution 103. The five squares are :

Solution 104. The number is 169.

Solution 105. We count nine squares : 1, 9, 25, 49, 81, 121, 169, 225 and 289.

Solution 106. On April 1^st 2013, Kate lives on the 58^th floor.

Solution 107. An arrangement of darts is :

Solution 108. Here are 10 results :

Solution 109. The number is 974.

Solution 110. Here is a way of representing each number from 7 to 12 :

Solution 111. N = 0, S = 1, M = 6, B = 8. The value of PB is 48.

Solution 112. The horizontal equalities are : 8 ´ 2 + 4 = 20, 3 + 7 – 1 = 9 and 8 – 5 ´ 6 = 18.

Solution 113. The number is 2520.

Solution 114. The ground measures 900 square meters. Its width is of 20 meters and its length of 45 meters.

Solution 115. An arrangement of balls is :

Solution 116. a = 4, b = 8, c = 9, d = 6, e = 5, f = 1 and g = 3. The phone number is 489-6513.

Solution 117. The number of passengers in each row is 78. The passengers are distributed thus :

Solution 118. The greatest expression is 8⁹.

Solution 119. The smallest result is 20. The operations are in order : (¸, -, ´, +) or (+, ¸, -, ´).

Solution 120. H = 1, E = 2, M = 3, J = 4, D = 5, A = 6, G = 7, P = 8 and B = 9. The value of JP is 48.

Solution 121. There are 14 possible paths.

Solution 122. We count the number of balls for each digit.

The number is 87.

Solution 123. The three-number combinations that have a sum of 21 are : (3, 7, 11), (3, 8, 10), (4, 6, 11), (4, 7, 10), (4, 8, 9), (5, 6, 10), (5, 7, 9) and (6, 7, 8). An arrangement is :

Solution 124. The numbers of the third circle are increased by 12 in comparison to the first. It is the same thing for the second and fourth circle. The numbers of the fourth circle are :

Solution 125. The total cost of the jacket and the shirt is 72 €. The cost of the shoes is 36 €. The cost of each item appears in this board.

Solution 126. We can write : 24 + 6 + 81 + 3 + 7 = 121.

Solution 127. Here is a way of representing 100 in each case :

Solution 128. We can count 16 grounds.

Solution 129. The bag contained 45 balls.

Solution 130. An arrangement of the numbers is :

Solution 131. This number is 24.

Solution 132. The result is always 4.

Solution 133. Two operations are repeated. We add up 3 and we subtract 1 successively. Here is each term with its rank :

The odd rank is always smaller of 1 than the term. The 15^th term is 16.

Solution 134. Martin plants 39 flowers in the corners. He has yet 60 flowers to plant. We make : 2 ´ 39 + 60 = 138 and 138 ¸ 3 = 46. Each row contains 46 flowers. An arrangement is :

At the center of the sides, he will plant 17, 19 and 24 flowers.

Solution 135. Here are two rectangles :

Solution 136. A square always ends in 0, 1, 4, 5, 6 or 9. As the sum of two squares ends in one 4, the two squares end in 0 and 4 or 5 and 9. The numbers are 15 and 7.

Solution 137. Here is the distribution of the colors :

Solution 138. Here are the six lozenges :

Solution 139. Remy will receive four tockens 1, four tockens 2 and one tocken 5.

Solution 140. As the sum is 22 220, the sum of numbers of each column must be 20. The four digits are 1, 2, 8 and 9. The numbers are 9182, 2891, 1928 and 8219.

Solution 141. The number is 36.

Solution 142. MATH = 8546 and POWER = 17 092.

Solution 143. We obtain 12 parts.

Solution 144. A square ends always in 0, 1, 4, 5, 6 or 9. Thus 547 602 is not a square.

Solution 145. An arrangement of the balls is :

Solution 146. The equality is : 841 + 148 = 989. RITE corresponds to 8419.

Solution 147. The number is 25.

Solution 148. K = 0 since J - J = K ; P = 1 since J ¸ J = P. As J + J and J ´ J are different, J cannot be 2. J cannot be greater than 3. Therefore J = 3, F = 6 and M = 9. FKMP is equal to 6091.

Solution 149. We have : 11 = 2² - 1 = 3, 111 = 2³ - 1 = 7 and 1 111 = 2⁴ - 1 = 15. We make : 3 ´ 7 ´ 15 = 315. There are 315 inhabitants in this village.

Solution 150. Each row must contain 16 points. We reverse the third and the fifth domino.

Solutions 1 to 50

Solutions 51 to 100