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Interesting second-grade Olympic competition
The fun second-year Olympiad 1 Olympiad is a rational spirit, which can make human thinking apply to the most perfect degree. Let's read the interesting second-grade Olympiad together-drink milk and feel the strange world of Olympiad!

Example 5: Xinxin drank a glass of milk. For the first time, she drank a third of the milk and filled it with water. The second time, she drank a third of the glass and filled it with water. After that, he finished this cup. Think about whether Xinxin drinks more milk or water.

Hugging: I want to know whether there is more milk or water. I can't count how much milk and water she drinks at a time, but how much milk she drinks and how much water she drinks as a whole. The milk was just a cup at first, without adding it. Add water three times, each time it is one third of a glass, that is, a glass of milk or water is divided into three parts and only one part is drunk, that is, only one part is added each time, and three times it is just one cup. I also drank a glass of water. It can be concluded that you drink as much milk as water.

Solution: Xinxin drinks as much milk as water.

Interesting second-year olympiad 2 1, Xiao Ming saw in the mirror that it was 5: 35 on the clock. Do you know what time it is?

2. Six ducklings line up, one in three teams, with three ducklings in each team. How to line up?

3, 8 people eat, each person 1 rice bowl, two people a dish bowl, four people 1 soup bowl, a * * *, how many bowls are there altogether?

4. Xiao Liang is sitting on the tour bus on the circular runway. He found six cars in front and six cars in the back. Excuse me: How many cars are there on the runway?

Suppose there is a plant that grows twice as tall every day. 20 days is exactly 20 centimeters high. Excuse me: When did it grow to 5 cm?

Xiaoming keeps some ducks at home. Know how many ducks there are. Think about it. "Half of the ducks are in the water, and half of them will go into the water if they are divided by 2, and the rest 15 will eat sundries around Xiao Ming. How many do you say? "

7, a division problem, the divisor is 6. Xiao Ming reversed the ten digits and single digits of the dividend, and the quotient of the dividend was 4. What is the correct quotient?

8, a bottle of oil, even 800 grams, eat half of the oil, even the bottle, there are 550 grams left. How many grams of oil are there in the bottle How many grams does this empty bottle weigh?

9 1 weight of goose+weight of 3 chickens =1weight of 0 ducks.

Weight of 8 chickens =1weight of 6 ducks.

1 weight of a goose = () weight of a duck.

1 goose = () the weight of a chicken.

10, 1 watermelon +2 pears = 16 apples.

5 pears = 10 apples 1 watermelon = () apples

1 watermelon = () pear

In the third week of the second year of the Olympics, the teacher sent a card to Xiaolan, Xiaoqing, Xiaohong, Xiaoling and Xiaohua in turn. The first card is sent to Xiaolan, the second card is sent to Xiaoqing, the third card is sent to Xiaohong, the fourth card is sent to Xiaoling, the fifth card is sent to Xiaohua, and the sixth card is sent to Xiaolan ... Miss Zhou * * * 40 cards. Who is the 23rd card addressed to? Who was the last letter addressed to?

Bid: 40 cards are sent to Xiaolan, Xiaoqing, Xiaohong, Xiaoling and Xiaohua in turn, that is, every 5 cards are a cycle, 235 = 4...3, and there are still 3 remaining cards. The first card must be sent to Xiaolan, the second card to Xiaoqing and the third card to Xiaohong according to the law.

Because MISS ZHOU has 40 cards, and the 40th card is the last one. 405=8, divisible, without remainder, which means that it is just the end of a cycle, and it must have been sent to the last floret among five people.

Answer: 235 = 4...3, the 23rd photo was sent to Xiaohong. 405=8, the last one is given to Xiaohua.

A: The 23rd photo was sent to Xiaohong, and the last photo was sent to Xiaohua.

Interesting second-year Olympiad 4 Olympiad is a rational spirit, which can make human thinking apply to the most perfect degree. Let's read interesting second-year Olympiad exercises together and feel the strange world of Olympiad!

Example 2: Write "Olympiad, Olympiad, Olympiad, Olympiad ……" in this order. What is the combination of the 19 word and the 26th word?

Hugging: The question takes every six words of "Mathematical Olympics" as a cycle and repeats them in turn. We have to figure out what the word 19 is first, then what the 26th word is, and then put these two words together to work out the answer.

Solution: 196 = 3... 1, the19th word is a number.

266 = 4 ...2. The 26th word is Xue.

A: Together, it is "mathematics".

The interesting second-year Olympic math exercises provided for you are combined together, hoping to bring you inspiration!

Interesting sophomore Olympiad 5 1. Solution: Classified calculation, using the number "1" to enumerate the numbers.

The number of "1" appearing in the cell is:

1, 1 1,2 1,3 1,4 1,5 1,6 1,7 1,8 1,9 1,

10 1, 1 1 1, 12 1, 13 1, 14 1, 15 1, 16 1, 17 1, 18 1, 19 1

***20;

"1" appears in the tenth digit:

10, 1 1, 12, 13, 14, 15, 16, 17, 18, 19

1 10, 1 1 1, 1 12, 1 13, 1 14, 1 15, 1 16, 1 17, 1 18, 1 19

***20;

"1" appears in hundreds of digits:

100, 10 1, 102, 103, 104, 105, 106, 107, 108, 109,

1 10, 1 1 1, 1 12, 1 13, 1 14, 1 15, 1 16, 1 17, 1 18, 1 19,

120, 12 1, 122, 123, 124, 125, 126, 127, 128, 129,

130, 13 1, 132, 133, 134, 135, 136, 137, 138, 139,

140, 14 1, 142, 143, 144, 145, 146, 147, 148, 149,

150, 15 1, 152, 153, 154, 155, 156, 157, 158, 159,

160, 16 1, 162, 163, 164, 165, 166, 167, 168, 169,

170, 17 1, 172, 173, 174, 175, 176, 177, 178, 179,

180, 18 1, 182, 183, 184, 185, 186, 187, 188, 189,

190, 19 1, 192, 193, 194, 195, 196, 197, 198, 199

*** 100;

The total number of times the number "1" appears in 1 to 200 is:

20+20+ 100= 140 (times).

2. Solution: enumeration method and classified calculation;

"3" in the unit: 3, 13, 23, 33, 43, 53, 63, 73, 83, 93 * *10;

Tenth "3": 31,33, 35, 37, 39 * * * 5;

From 1 to 100, the total number of times the number "3" appears in odd numbers:

10+5= 15 (times).

3. Solution: enumeration method: 12, 17, 22, 27, 32, 37, 42, 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, 97 * * 67.

4. Solution: Count by sections and then add up.

Number of pages and types

1 ~ 9 * * * 9 pages 1×9=9 (sheets) (use 1 type for each page number)

10 ~ 90 * * * 90 pages 2×90= 180 (sheets) (use 2 types for each page number)

100 ~199 * *100 Page 3× 100=300 (sheets) (use 3 pages).

Page 200 *** 1 Page 3× 1=3 (pieces) (this page uses 3 types)

Total: 9+ 180+300+3=492.

5. Solution: list enumeration, classified statistics:

10 1

202 12

303 1323

404 142434

505 15253545

606 1626364656

707 172737475767

808 18283848586878

909 192939495997989

The total is 1+2+3+4+5+6+7+8+9=45.

6. Solution: Enumeration, and then total:

10 1, 1 1 1, 12 1, 13 1, 14 1,65438+