There is a box with chocolates inside it. When we count the chocolates in pairs, one is left. When we count the chocolates in a group of three, two chocolates are left. When we count the chocolates in a group of four, three chocolates are left. When we count the chocolates in a group of five, four are left. When we count the chocolates in a group of six, five are left.
But when the chocolates are counted in a group of seven, no chocolate is left. Can you calculate the minimum number of chocolates that are present in the box?	Options:
A) 113
B) 116
C) 119
D) 118
E) 112
C   119   Let the minimum number of chocolates = C   When divided by 2, remainder = 1   When divided by 3, remainder = 2   When divided by 4, remainder = 3   When divided by 5, remainder = 4   When divided by 6, remainder = 5   When divided by 7, remainder = 0   Therefore, C is divisible by 7.   Whenever C is divided by any number less than 7, the remainder is 1 less than the divisor.   This implies that, C + 1 is the LCM of 2, 3, 4, 5 and 6.   Now L119M of 2, 3, 4, 5 and 6 = 60   But  60 - 1= 
<gadget id="calculator"> 60 - 1</gadget>
<output>59</output>
59 is not divisible by 7   60 * 2= 
<gadget id="calculator">60 * 2</gadget>
<output>120</output>
120   120 - 1= 
<gadget id="calculator">120 - 1</gadget>
<output>119</output>
119 which is divisible by 7   Therefore, 119 + 1= 120 or 119 = 119   Number of minimum number of chocolates in the box = 119
<result>C</result>
There is a box with chocolates inside it. When we count the chocolates in pairs, one is left. When we count the chocolates in a group of three, two chocolates are left. When we count the chocolates in a group of four, three chocolates are left. When we count the chocolates in a group of five, four are left. When we count the chocolates in a group of six, five are left.
But when the chocolates are counted in a group of seven, no chocolate is left. Can you calculate the minimum number of chocolates that are present in the box?