一个数字乘以2得到60。那个数字是多少?该数字是30。该数字未知。假定该数字为变量,并为该变量分配什么值,以得到结果为60。变量定义为未知数量。没有确定的值,并且一直在变化。这是一种难题解决类型,其中需要确定指定字母的值。字母的值必须只有一位数字。如果拼图中有多个字母,则不能将相同的值分配给多个字母。如果字母处于起始位置,则字母的值不能为零。
数字字母
基于此主题的问题是一些难题,其中字母在算术加法中代替了数字,而问题是要找出哪个字母代表哪个数字。这种令人费解的问题似乎就像是使用算术运算来破解代码。
在这种类型的谜题中,字母代表数字,并且需要破解代码,记住一些要记住的事情,例如:
- Each digit in the arithmetic operation represents a single letter and vice-versa.
- Any number used in this, can not have first digit as zero e.g. 25 should not be 025 or 0025.
- And every puzzle must have only one solution or code.
益智与加法
考虑以下示例,该示例具有一个加法拼图,如下所示:
Example: Find the value of unknown Y in the following addition:
Solution:
It is known that for the given addition there is only one letter Y whose value is needed to determine.
Consider the addition of the first column i.e. from Y + 3 which gives 1 at its one’s place. This implies that Y can be 8 because 8 + 3 = 11 i.e. 1 at one’s place.
If this is so then the given addition must be written as:
Hence, the value of the unknown Y is 8.
乘法益智
考虑下面的例子,下面给出了一个带有乘法的难题:
Example: Determine the values of the unknowns P and Q in the following multiplication.
Solution:
Consider the multiplication from the first column i.e. Q × 7 which again gives Q at its one’s place. This implies that Q can be either 0 or 5 because only 0 × 7 = 0 itself and 5 × 7 = 25.
Now, in order to determine the value of P following are the cases:
When P = 1 and Q = 0 then A B × A 7 = 1 0 × 1 7 = 170,
When P = 1 and Q = 5 then A B × A 7 = 1 5 × 1 7 = 255,
When P = 2 and Q = 0 then A B × A 7 = 2 0 × 2 7 = 540,
When P = 2 and Q = 5 then A B × A 7 = 2 5 × 2 7 = 675,
When P = 3 and Q = 0 then A B × A 7 = 3 0 × 3 7 = 1110 and
When P = 3 and Q = 5 then A B × A 7 = 3 0 × 3 7 = 1295.
Hence, the required values of the unknown is: P = 3 and Q = 5.
样本问题
以下是基于上述主题的一些问题:
问题1:找到未知数C和D的值:
解决方案:
Consider the addition of the first column i.e. from C + 5 which gives 4 at its one’s place. This implies that C can be 9 because 9 + 5 = 14 i.e. 1 at one’s place.
If this is so then the given addition must be written as:
Hence, the value of the unknowns C = 9 & D = 8.
问题2:在以下加法运算中将Z的值替换为真:
解决方案:
Firstly, consider the addition of the first column i.e 7 + 5 = 12 this implies that the addition in second column is: 1 + Z + 8 to give 5 at one’s place. For this case the Z should be 6.
If this is so then the given addition must be written as:
Hence, the value of Z is equal to 6.
问题3:在下面的乘法问题中,用P和Q的值替换为真:
解决方案:
For the given case the product of Q and 6 must be same as ones place digit i.e Q.
Now,
1 × 6 = 6,
2 × 6 = 12,
3 × 6 = 18,
4 × 6 = 24
Thus, one of the digit multiplied and the ones place digit of the product obtained is same.
So, the given multiplication becomes:
Now, the value of P is obtained as:
On putting Q = 4, the ones digit must be obtained as 4 and the remaining two values of Q should be 44, which can be possible when 7 is multiplied with 6 to give 44. Therefore, the given multiplication becomes:
Hence, the value of P and Q are 7 and 4.
问题4:在以下加法问题中,X和Y的值替换为真:
解决方案:
Consider the first column of the addition i.e. the sum of X and 5 gives 9 at one’s place, this can be possible when 4 is added to 5.
Therefore, the given addition problem becomes:
Hence, the value of X and Y are 4 and 9.
问题5:当A 3 + 8 B = 150时,A – B的值是多少?
解决方案:
Consider the addition of one’s place digits first i.e. the sum of 3 and B is such that the digit at one’s place is 0.
This is possible when B is 7 i.e. 3 + 7 = 10. So the given addition becomes: 6 3 + 8 7 = 150.
This implies that A = 6 and B = 7.
Thus, A – B = 6 – 7 = -1.