[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 512
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By the way, I found a beauty for 5.
So, we have 3 odd digits (1 3 5) which will give a 5 when multiplied by 5.
And since hockey digits are only 123456, only two (5 6) >= 5, i.e. one 5 must be converted to one (at least), which is unrealistic.
Hurrah, comrades, now we can calm down and quietly finish the file libka.
Assemble the solution in its entirety. If there is any divisibility, it is only by integers in the range from 2 to 5.
Simulate multiplication in columns with memorization and transferring what is "in mind" to a higher digit.
TheXpert: На 2 стопудофф нельзя умножать, т.к. в результате получается максимум 2 нечетные цифры, а надо 3. Ну и 2*4 = 8
Ha, you can't do 3 because 3*6 = 8, no way to get 1...6.
You can't do 4 because 2*4 = 8, 6*4 = 24, there's no way to get 1 from 8.
That leaves us with 5.
TheXpert: So, we have 3 odd digits (1 3 5) which give 5 when multiplied by 5.
And since hockey digits are only 123456, only two (5 6) >= 5, i.e. one 5 must be converted to one (at least), which is unrealistic.
The explanation for multiplier 2 is more like this ("at most 2 odd digits" is a lot of overkill, where a lot depends on the mutual arrangement of the digits):
Avals: The digit 2 as a multiplier is not suitable. If we multiply by columns as at school))), in that position of hockey number where 4 when multiplied will be 8, and in order to reach 1 (also a hockey number), then in mind d.b. 3 - i.e. in the previous digit when multiplied should turn out more than thirty, and this is impossible with the given multiplier and hockey numbers
It's all worked out great. The whole world piled in, and even wrote a program. Here's another problem for those for whom file libs are not a top priority:
Pupils of a mathematical class stand in a row (there are both girls and boys in the class).
It is known that any two pupils with exactly 12 or exactly 19 other pupils standing between them are of the same gender.
a) Find the largest possible number of students in the class.
b) How would the answer to the problem change, if you replace "in row" by "in circle"?
And here is the solution to the hockey players problem given by the girl who posted it:
Сумма цифр каждого хоккейного числа равна 21, сиречь, даёт остаток 3 при делении на 9.
Стало быть, если одно хоккейное число делится на другое, их отношение может быть только 4 или 7, но 7 отпадает, ибо тогда большее число не меньше 700000.
Значит, только 4.
А теперь присмотритесь внимательно, что происходит с двойкой.
Если записать четыре двойки одна под другой, выйдет 8, 9 или 0.
Больше выйти не может, ибо тогда придётся занимать из предыдущего разряда как минимум тройку, что, очевидно, невозможно.
The answer is that there are no such numbers.
By the way, there was already a comment about the sum of the numbers. It just wasn't noticed.
So it is only 4.
What does dividing by 9 have to do with it? And how does marking all the divisors except 4 and 7 follow from the remainder?
And here is the solution to the hockey players' problem given by the girl who posted it:
1. The sum of the digits of each hockey number is 21, which gives a remainder of 3 when divided by 9.
2. Hence, if one hockey number is divided by another, their ratio can only be 4 or 7,
By the way, there's already been an observation about the sum of the numbers. It just hasn't been noticed.
What does division by 9 have to do with it? And how does marking all divisors except 4 and 7 follow from the remainder?
The theory of modulo comparisons is a very powerful thing.
The sum of the digits of any hoc number is always 21 = 3(mod 9). By the rule of divisibility by 9, it follows that any hockey number also has a remainder of 3 when divided by 9. Consequently, n*HockeyNumber = n*3 (mod 9).
Multiplying a hockey one by 2 will make the remainder of mod 9 equal to 6 - i.e. the number becomes a non-hockey one.
Multiplying by 3 makes the number a multiple of 9 - also non-hockey.
Multiplying by 4: 4*3 (mod 9) = 3 (mod 9) - possibly hockey.
By 5: 4*5 (mod 9) = 6 (mod 9) - not hockey.
You don't need to check any further.
Pupils in a maths class stand in a row (there are both girls and boys in the class).
We know that any two pupils with exactly 12 or exactly 19 other pupils standing between them are of the same gender.
a) Find the largest possible number of pupils in the class.
b) How would the answer to the problem change if "in row" was replaced by "in circle"?
For a, I got 29: If M=1, D=0 then
11100001110001110000111000111
B.F. for b it seems to be 3 less (26) because the construction for a does not fit the last three units
Modulo comparison theory is a very powerful thing.
for a I got 29: if M=1, D=0 then
11100001110001110000111000111
B.F. for b it seems to be 3 less (26) because the construction for a does not fit the last three units
The theory of modulo comparisons is a very powerful thing.
The sum of the digits of any hockey number is always 21 = 3(mod 9). According to the divisibility by 9, it follows that any hockey number also has a remainder of 3 when divided by 9. Consequently, n*HockeyNumber = n*3 (mod 9).
Multiplying a hockey one by 2 will make the remainder of mod 9 equal to 6 - i.e. the number becomes a non-hockey one.
Multiplying by 3 makes the number a multiple of 9 - also non-hockey.
Multiplying by 4: 4*3 (mod 9) = 3 (mod 9) - possibly hockey.
By 5: 4*5 (mod 9) = 6 (mod 9) - not hockey.
You don't need to check any further.