Urgently, to ensure that it is right, urgent

If 10 are in a bag, there is one more at the end: 9 in a bag, and at the end there is one more: 8 in a bag, and one more at the end. 2 in a bag, and in the end there is still one more.

Seek 10, 9, 8, 7....2,1 is the least common multiple.

The number of apples in this pile = 2521 + 2520n (n is the natural number) because there are about 5000 5100 apples in a pile.

So n = 1, then the number of apples in this pile = 2519 + 2520 = 5041.

It is common sense to find 10, 9....a common multiple of 3,2, however, because 10 already contains 2,5; 9 already contains 3; 8 contains 4; The factor of 6 is also contained by 9 and 8, respectively; So the common multiples of 9,8,7,5 should be found).

As can be seen from the title, the number of apples in this pile minus 1 is a common multiple of 10,9,8,2, and is between 5000-5100.

Then the common multiple is 5040 and the total number of apples is 5041

5041 pcs.

Because the number of apples in this pile is a common multiple of 10, 9, 8, 7, 6, 5, 4, 3, 2, plus 1, you can first find the common multiple of 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, .

The number of this requirement must be n times of 10,9,8,7,6,5,4,3,2,1 of the least common multiple, after calculating their least common multiple is 2520, it is known that the range of this number is 5000-5100, and the value of n can be known to be 2, then you can find the number of this pile of apples is 5041 (I believe you will ask for this common multiple, I will not explain in detail.)

Length of shared ribbon: (6*6+ cm.)

6000 loss is to be staggered to jujube 15=400

6000+400 is 6400

6000 12 is 500

6400+500 is 6900

The answer is 6900

altogether = 5000 (1-4 10-4 8).

50000 pcs.

There is a brigade to help the ants to help remember the praise, new questions, please re-post to mention the dismantling of the cover to ask, no longer thank you here.

If you dig on the side: 6 * 10 * 10 - 2 * 2 - 2 * 2 = 592 square centimeters.

If you don't dig on the side: 6 * 10 * 10 + 10 * 2-2 * 2 - 2 * 2 = 612 square centimeters.

Situation 1: Dig against the surface.

10 x 10 x 6 - 2 x 2 x 2 + 2 x 10 x 2

Case 2: Dig in the middle.

10 x 10 x 6 - 2 x 2 x 2 + 2 x 10 x 4

Cube surface area 10 x 10 x 6

As the title says, because the cuboid is 10cm long

There are 4 more recessed large sides and 2 less small sides.

But case 1 is a special case, with only two more concave large sides and two less small sides.

10*10*6 (surface area of the original cube) -2*2*2 (digging out the small cuboid actually only reduces the surface area of the height and bottom of the sand body) = 592