Numbers, especially integers, are the earliest and most widely known numbers. In the kingdom of integers, there are treasures left by predecessors waiting for us to collect, and there are fascinating questions waiting for us to explore. This is a fascinating and beautiful world, a world where we can gallop freely.
Learning mathematics, of course, can not be separated from calculation, students must hope that they can calculate correctly and quickly, so how can we do this?
First of all, we must master the calculation rules and operation order skillfully; Secondly, we should choose a reasonable and flexible calculation method according to the characteristics of the topic itself.
For example, calculate the following question:
( 1)28+49+72+5 1; (2)763-278-322;
(3) 125×56; (4)4500÷25÷4.
The above four calculation problems are very simple, and I believe students will work out the correct results. But how did you do it? Is it possible to simplify the calculation?
When calculating, I think all the students have this experience: it is much faster to make whole ten, whole hundred, whole thousand and ... In fact, from this basic experience, students can extract a very common fast calculation method-"rounding method".
Observing the above formula, it is not difficult to find that the sum of 28 and 72, 49 and 5 1 in (1) can be added to 100, and the sum of 278 and 322 in (2) is 600. Grasping this feature, we can mentally calculate that the results of these two questions are 200 and 65438 respectively. Question (4) can be changed to 4500 (25× 4), then we can quickly get the results of questions (3) and (4) are 7000 and 45 respectively.
Question 1. 1 Calculate the following question:
( 1)729+54+27 1;
(2) 136 1+972+639+28;
(3) 12345+4680 1+87362+87655+53 199+ 12638.
solve
( 1)729+54+27 1=(729+27 1)+54
= 1000+54= 1054;
(2) 136 1+972+639+28=( 136 1+639)+(972+28)
=2000+ 1000=3000;
(3) The original formula = (12345+87655)+(46801+53199)+(87362+12638).
= 100000+ 100000+ 100000=300000.
From the answer to the above question 1. 1, we can see that when calculating the sum of several addends, we can use the exchange law and associative law of addition to add two numbers that can be rounded first, and then add the obtained sum, which can greatly simplify the calculation.
Question 1.2 Calculate the following question:
( 1)66+75+38;
(2)9998+3+99+998+3+9;
(3) 19999+ 1999+ 199+ 19+9.
Analyze and observe the characteristics of this group of questions. Compared with the question 1. 1, each question in the question 1.2 does not directly give two integers, but we can decompose one of the addends into the sum of two numbers (or add a number), so that one of the numbers can be rounded with one of the addends of the question, and the sum of the two can be added.
In (1), look at 66 and decompose 38 into the sum of 34 and 4; In (2), look at 9998, 998, 99 and 9, and decompose two 3s into the sum of 2 and 1; In (3), look at 19999, 1999, 19, decompose 9 into the sum of 5 and 4 1, or add 5 1, and the problem will be solved.
solve
( 1)66+75+38=(66+34)+(75+4)
= 100+79= 179;
(2)9998+3+99+998+3+9
=(9998+2)+( 1+99)+(998+2)+( 1+9)
= 10000+ 100+ 1000+ 10= 1 1 1 10;
(3) 19999+ 1999+ 199+ 19+9
=( 19999+ 1)+( 1999+ 1)+( 19+ 1)+( 19+ 1)+5
=20000+2000+200+20+5=22225.
Question (3) can also be calculated as follows:
19999+ 1999+ 199+ 19+9
=( 19999+ 1)+( 1999+ 1)+( 199+ 1)+( 19+ 1)+(9+ 1)-5
=20000+2000+200+20+ 10-5=22225.
Question 1.3 Calculate the following question:
( 1)76543+498; (2)9999+999+99+9;
(3) 1238+2759-98-997;
(4)27.6+ 16.5+72.4+ 18.7+43.5.
Students can simply calculate the results of each question in 1.3 by using the "rounding method" learned above, but for (4), "rounding" does not need to be rounded to whole ten, whole hundred, whole thousand, ..., but only to whole numbers.