Math exercises and answers in the sixth grade of primary school
Example 1. Only by modifying a number 970405 can the modified six digits be divisible by 225, and the modified six digits are _ _ _ _ _. (Anhui province 1997 primary school mathematics competition) solution: reverse thinking: because 225=25×9, and 25 and 9 are coprime. Let's examine the divisibility of 25 and 9 respectively. According to the feature that the number can be divisible by 25 (the last two digits can be divisible by 25), the last two digits of the modified six-digit number may be 25 or 75. According to the characteristics of numbers divisible by 9 (the sum of digits in each digit can be divisible by 9), it is 9+7+0+4+5 = 25, 25+2 = 27, 25. Answer 48 There are three choices for solving hundreds, namely 1, 4, 9, and four choices for tens and digits, namely 0, 1, 4, 9. There are 3× 4× 4 = 48 three-digit * * to satisfy the meaning of the question. 12. The product of each digit of a given three-digit number is equal to 10, so the number of three digits is _ _ _ _. Answer 6 Because 10 = 2× 5, these three digits can only be composed of 1, 2,5, so * * has = 6. The sum of three numbers in each triangle is equal to 50, where A7 = 25, A 1+A2+A3+A4 = 74, and A9+A3+A5+A 10 = 76. What is the sum of A2 and A5? Answer 25 The solution is A 1+A2+A8 = 50, A9+A2+A3 = 50, A4+A3+A5 = 50, A 10+A5+A6 = 50, A7+A8+A6 = 50, So a1+A2+A8+A9+A2+A3+A4+A3+A5+A1A5+A6+A7+A8+A6 = 250, that is, (a 654438 has 74+76+A2+A5+. So A6+A8 = 50-25 = 25. Then A2+A5 = 250-74-76-50-25 = 25. It is suggested that the above deduction is completely correct, but we lack a sense of direction and overall grasp. In fact, when we see such a digital array, our first feeling is that the five 50s here do not refer to the sum of 10 numbers, but the sum of this 10 number plus the number of the inner circle 5. This is the most obvious feeling and an important equal relationship. Then "look at the problem and set the direction", find the sum of the second number and the fifth number, which means it is related to the other three numbers in the inner circle. The sum of the sixth number and the eighth number is 50-25 = 25, and then look at the third number. When the numbers 1, 2, 3 and 4 of two straight lines are added with the numbers 9, 3, 5 and 10, the drama begins: 74+76+50+25+ the second number+the fifth number = 50 × 5, so the second number+the fifth number = 25. Fill in the blanks first: 1 satisfies the following formula. * * * What kind? Word of mouth-Word of mouth = Word of mouth answer 4905. You can know the answer from the correct formula. This problem is equivalent to how many formulas are there to find out that the sum of two digits A and B is not less than 100. When a= 10, b is between 90 and 99, and there are 10 species; When a= 1 1, b is between 89 and 99, with 1 1 species; ..... when a = 99, there are 99 kinds of B between 1 99. * * * You have10+1+12+... 99 = 4905 (species). It is suggested to combine the mystery of formula with the counting problem, which is an example. The analogy association of mathematical models is the key to solving problems. There are pentagons and hexagons on the football surface (see the top right). Each pentagon connects five hexagons, and each hexagon connects three pentagons. Then the simplest integer ratio of pentagon and hexagon is _ _ _ _ _. The answer is 3: 5. This solution has x pentagons. Each pentagon is connected with five hexagons, so there should be 5X hexagons, but each hexagon is connected with three pentagons, that is, each hexagon counts three times, so there are six hexagons. Second, answer the question: 1. Xiaohong went to the store and bought a box of flowers and a box of white balls. The number of two boxes of balls is equal. The original price of flower balls is 3 in 2 yuan and 5 in 2 yuan. There was a discount on the New Year's Eve. The price of both balls was 4 yuan and 8 pieces. As a result, Xiaohong spent 5 yuan less. So, how many balls did she buy? The answer 150 is analyzed with a rectangular diagram, as shown in the figure. Easy to get, easy to solve: